Discussion:
Real amp draw of E3000
(too old to reply)
William Enestvedt
2006-09-21 13:38:59 UTC
Permalink
Can anyone suggest what load a Sun Enterprise 3000 really draws?

The Sun docs and System Handbook web pages say it requires a 15 amp
circuit (duh) but I can't find an actual figure on what the maximum or
average figure are.

I'm re-racking a few things in my data center, and I would hate to
overload the PDU's circuit breakers. :7)

Thank you in advance.

- Will
P.S. I am a literature major and I learned all my home improvement
skillz from "This Old House," so if "load" and "draw" are the wrong
terms, feel entirely free to be as abusive as you like when correcting
me -- just make sure to provide an answer. :7)
--
Will Enestvedt
UNIX System Administrator
Providence, RI
Patrick Giagnocavo
2006-09-21 13:46:05 UTC
Permalink
Post by William Enestvedt
Can anyone suggest what load a Sun Enterprise 3000 really draws?
The Sun docs and System Handbook web pages say it requires a 15 amp
circuit (duh) but I can't find an actual figure on what the maximum or
average figure are.
I think, based on my experience with an E4000, that you should budget
750W available for when it starts, and about 400W to 500W continuous. I
think 340W continuous was what the E4000 used with 8x250Mhz CPUs, 5-6GB
and a few internal disks.

This is all from memory, though.

--Patrick
Jonathan C. Patschke
2006-09-21 15:39:40 UTC
Permalink
Post by Patrick Giagnocavo
I think, based on my experience with an E4000, that you should budget
750W available for when it starts, and about 400W to 500W continuous.
I think 340W continuous was what the E4000 used with 8x250Mhz CPUs,
5-6GB and a few internal disks.
My E4000 pulled substantially more power at 10 336MHz CPUs, 10GB memory,
and two I/O boards. IIRC it was closer to a kilowatt -sustained- under
maximum CPU load. It was a staggering amount of power for having no
disks whatsoever, but it was a very dense package of logic.

My E3000 doesn't pull much power. It's configured with 4.5GB memory,
six 250MHz processors, and four 10krpm disks. It's a lot closer to the
500W continuous figure you gave.
--
Jonathan Patschke "The ruler demands gifts, the judge accepts bribes,
Elgin, TX the powerful dictate what they desire--they all con-
USA spire together. The best of them is like a brier, the
most upright worse than a thorn hedge." --Micah 7:3-4
Phil Stracchino
2006-09-21 15:42:40 UTC
Permalink
Post by Jonathan C. Patschke
Post by Patrick Giagnocavo
I think, based on my experience with an E4000, that you should budget
750W available for when it starts, and about 400W to 500W continuous.
I think 340W continuous was what the E4000 used with 8x250Mhz CPUs,
5-6GB and a few internal disks.
My E4000 pulled substantially more power at 10 336MHz CPUs, 10GB memory,
and two I/O boards. IIRC it was closer to a kilowatt -sustained- under
maximum CPU load. It was a staggering amount of power for having no
disks whatsoever, but it was a very dense package of logic.
My E3000 doesn't pull much power. It's configured with 4.5GB memory,
six 250MHz processors, and four 10krpm disks. It's a lot closer to the
500W continuous figure you gave.
Mine was 6x336, 4GB RAM, and 10x10KRPM. I never actually measured the
exact draw, though, I just estimated based on what else I could have on
the circuit with it without tripping the breaker.
--
Same geek, same site, new location
Phil Stracchino Landline: 603-429-0220
***@speakeasy.net Mobile: 603-216-7037
Renaissance Man, Unix generalist, Perl hacker, Free Stater
William Enestvedt
2006-09-21 13:54:34 UTC
Permalink
Post by Patrick Giagnocavo
I think, based on my experience with an E4000, that you should budget
750W available for when it starts, and about 400W to 500W continuous.
I think 340W continuous was what the E4000 used with 8x250Mh
CPUs, 5-6GB and a few internal disks.
I *love* real world data, thank you!

When I called Sun last year about the PSUs on a V490 with 208 plugs,
I got a long run-around saying that I would have to _hire_ Sun
professional Services in order to ask this question. See, the Sun
Spectrum Support people couldn't tell me anything about the power as it
was "a liability issue." *snort*

Have you got any idea about the amps, though? I think my twin PDUs
are rated for 20 amp (on 20 amp circuits, I know that much!), and I want
to put in a V40z (10 amp max.) where I already have a 420R (6.3), a
4-CPU E3500 (9.5), a 280R (9.2), a 12-drive A1000 (2.6), and a 15" CRT.

I know, I know, that's a total of almost 30 amps now with the V40z
bringing in up to 10 more -- but it's never failed before, and I really
don't think I can trust those maximums for real world math.

*sigh* Maybe I should have stayed in engineering school instead of
getting an English degree...

Thanks again.

- Will
--
Will Enestvedt
UNIX System Administrator
Providence, RI
Will McDonald
2006-09-21 15:26:53 UTC
Permalink
Post by William Enestvedt
Post by Patrick Giagnocavo
I think, based on my experience with an E4000, that you should budget
750W available for when it starts, and about 400W to 500W continuous.
I think 340W continuous was what the E4000 used with 8x250Mh
CPUs, 5-6GB and a few internal disks.
I *love* real world data, thank you!
When I called Sun last year about the PSUs on a V490 with 208 plugs,
I got a long run-around saying that I would have to _hire_ Sun
professional Services in order to ask this question. See, the Sun
Spectrum Support people couldn't tell me anything about the power as it
was "a liability issue." *snort*
Have you got any idea about the amps, though? I think my twin PDUs
are rated for 20 amp (on 20 amp circuits, I know that much!), and I want
to put in a V40z (10 amp max.) where I already have a 420R (6.3), a
4-CPU E3500 (9.5), a 280R (9.2), a 12-drive A1000 (2.6), and a 15" CRT.
http://www.simetric.co.uk/si_watts.htm

A = W / V

So:

A = 750/240 (for the UK)
A = 3.125 (again for the UK)

A = 750/120 (for the US?)
A = 6.25 (again for the US?)


nb: I'm not qualified in engineering either, therefore for liability
reasons I recommend you hire an electrical engineer to perform these
calculations for you. ;)

Will.
Mike Meredith
2006-09-21 17:11:12 UTC
Permalink
Post by William Enestvedt
I *love* real world data, thank you!
Not sure about real world data but some real world experience ... not so
far back I watched half a 'data centre' loose power twice in 5 hours
because of loading issues (and a *very* big burn out fuse).
Post by William Enestvedt
Have you got any idea about the amps, though? I think my twin PDUs
are rated for 20 amp (on 20 amp circuits, I know that much!), and I
want to put in a V40z (10 amp max.) where I already have a 420R (6.3),
a 4-CPU E3500 (9.5), a 280R (9.2), a 12-drive A1000 (2.6), and a 15"
CRT.
Amp? Isn't that what you use to make sounds bigger ?

Systems have a specified load that's far bigger than they usually draw
... after all most of our systems (even the 10-CPU E4000s) aren't loaded
with as much power-sucking stuff as they could be (an E4000 probably
draws more power with 2 CPUs and loads of I/O boards with funky SBUS
cards, and a few disk trays).

So the fact that you're running systems totalling 30amps on a 20amp PDU
isn't too surprising. You may or may not be able to add the V40z to the
mix, however ...

Systems draw more power when first switched on I believe (and certainly
a qualified electrician explained what happened that way) so in the
unusual circumstances of powering on a whole PDU-worth of equipment at
once, you can well overload the PDU which works with the systems in
'steady state' mode.

After our burnt out fuse was replaced, we reset the PDU (somewhat bigger
than 20amps) and the above was exactly what happened to us ... the
starting load overloaded the PDU and it reset back to off. Ensuring that
half of the systems were off before resetting was somewhat tedious.

I think I was in for 20 hours that day.

It's safest to assume systems will pull their maximum specified load,
and work around that. More expensive too though.
Post by William Enestvedt
*sigh* Maybe I should have stayed in engineering school instead of
getting an English degree...
Well, maybe. It's worth noting that even the PHB in charge of
implementing our new data centre (in a previous life was regularly
handling much more electrickery than even a large data centre pulls) was
most insistent that the electrical requirements were handled by
currently qualified electricians.
Charles Shannon Hendrix
2006-09-21 22:11:01 UTC
Permalink
Post by Mike Meredith
Amp? Isn't that what you use to make sounds bigger ?
Only if it has an 11...
--
shannon "AT" widomaker.com -- ["Consulting wouldn't be what it is today
without Microsoft Windows" -- Chris Pinkham]
Brooke Gravitt
2006-09-21 13:54:38 UTC
Permalink
Post by William Enestvedt
P.S. I am a literature major and I learned all my home improvement
skillz from "This Old House," so if "load" and "draw" are the wrong
terms, feel entirely free to be as abusive as you like when correcting
me -- just make sure to provide an answer. :7)
I deference to "National Talk Like A Pirate Day" I think the appropriate
phraseology would be

"Yarr, I be stashin' me booty, me not be wantin' to keelhaul me olde
scrimshaw."

Or similar.
Lionel Peterson
2006-09-21 15:10:00 UTC
Permalink
Date: 2006/09/21 Thu AM 08:38:59 CDT
Subject: [rescue] Real amp draw of E3000
<snip>
P.S. I am a literature major and I learned all my home improvement
skillz from "This Old House," so if "load" and "draw" are the wrong
terms, feel entirely free to be as abusive as you like when correcting
me -- just make sure to provide an answer. :7)
Well, as long as it wasn't "Tool Time" on the "Home Improvement" TV show you should be OK ;^)

Lionel
Phil Stracchino
2006-09-21 15:29:38 UTC
Permalink
Post by William Enestvedt
Can anyone suggest what load a Sun Enterprise 3000 really draws?
The Sun docs and System Handbook web pages say it requires a 15 amp
circuit (duh) but I can't find an actual figure on what the maximum or
average figure are.
I'm re-racking a few things in my data center, and I would hate to
overload the PDU's circuit breakers. :7)
I think when I had my E3000 running, it was drawing about 8 amps after boot.
--
Same geek, same site, new location
Phil Stracchino Landline: 603-429-0220
***@speakeasy.net Mobile: 603-216-7037
Renaissance Man, Unix generalist, Perl hacker, Free Stater
William Enestvedt
2006-09-21 16:57:04 UTC
Permalink
Post by Will McDonald
A = W / V
A = 750/240 (for the UK)
A = 3.125 (again for the UK)
A = 750/120 (for the US?)
A = 6.25 (again for the US?)
"Aiiieeee! MATH!!!" *head 'splosion*
Post by Will McDonald
I think when I had my E3000 running, it was drawing about 8 amps after
boot.
Based on those two replies, I think I have about the same load in the
old full rack as the new one without the V40z, so I will look elsewhere
for available capacity.

Thanks to all to helped me avoid a st00pid mistake.

- Will
P.S. That A=W/V equation looks familiar. I just wish I could remember
that kind of stuff. :7( Mind you, for some reason, I haven't been able
to get PV=nRT out of my head since tenth grade, though I'm hazy on what
use it is to me now that I don't measure anything in moles or
atmospheres anymore...
--
Will Enestvedt
UNIX System Administrator
Providence, RI
der Mouse
2006-09-21 16:59:28 UTC
Permalink
Post by William Enestvedt
P.S. That A=W/V equation looks familiar.
It should; it's just a slight rearrangement of W = V W A.

However, it's good only for resistive loads or instantaneous
measurements; if your load is not purely resistive, your power consumed
is generally not your RMS volts times your RMS amps - it's the integral
of your instanteous volts times your instantaneous amps. (For example,
a purely capacitive or purely inductive load would draw current but
consume no power - not that such a thing actually exists in practice.)

/~\ The ASCII der Mouse
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Patrick Finnegan
2006-09-21 18:28:58 UTC
Permalink
Post by der Mouse
Post by William Enestvedt
P.S. That A=W/V equation looks familiar.
It should; it's just a slight rearrangement of W = V W A.
I'm gonna guess that's a typo, and you meant W = V x A.
Post by der Mouse
However, it's good only for resistive loads or instantaneous
measurements; if your load is not purely resistive, your power
consumed is generally not your RMS volts times your RMS amps - it's
the integral of your instanteous volts times your instantaneous amps.
(For example, a purely capacitive or purely inductive load would
draw current but consume no power - not that such a thing actually
exists in practice.)
A more common simplification is to say that the power used (in watts) is
the voltage times the amperage times the cosine of the phase angle
between the two (assuming your power is AC with a nice sine wave). A
purely inductive load (with a +90 deg phase angle) or purely capacitive
load (with a -90 deg phase angle) thus uses 0 watts, as cos(+-90deg) ==
0. This number is also called the "power factor" of the device, and
optimally should be very close to 1.


However, Volts x Amps always is equal to the VA (volt-amp) rating of a
device, and is useful for sizing breakers, or UPSs. Unlike Watts,
though, you can't generally directly add VA ratings, because they
aren't normalized to a particular phase angle. You have to take the
phase angle/power factor into account, which means yet more math....

Will, any chance you got any of that? :)

Pat
--
Purdue University Research Computing --- http://www.rcac.purdue.edu/
The Computer Refuge --- http://computer-refuge.org
Will McDonald
2006-09-21 19:25:29 UTC
Permalink
[BEER, BEER, BEER, BEER, BEER. BED, BED, BED, BED, BED.]
Will, any chance you got any of that? :)
I'm visualising Will like this right now:

Loading Image...

Will.
der Mouse
2006-09-21 19:43:19 UTC
Permalink
Post by Patrick Finnegan
Post by der Mouse
It should; it's just a slight rearrangement of W = V W A.
I'm gonna guess that's a typo, and you meant W = V x A.
Actually not a typo in the usual sense of the term. The W on the
right-hand side was, when I sent it, an ISO 8859-1 multiplication sign,
0xd7. Something somewhere between me and list output stripped the high
bit, turning it into 0x57, W.
Post by Patrick Finnegan
Post by der Mouse
if your load is not purely resistive, your power consumed is
generally not your RMS volts times your RMS amps - it's the integral
of your instanteous volts times your instantaneous amps.
A more common simplification is to say that the power used (in watts)
is the voltage times the amperage times the cosine of the phase angle
between the two (assuming your power is AC with a nice sine wave).
This works only if your current is also a nice sine wave. With a lot
of modern loads (in particular, anything with a switching power
supply), this is far from true.

/~\ The ASCII der Mouse
\ / Ribbon Campaign
X Against HTML ***@rodents.montreal.qc.ca
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Phil Stracchino
2006-09-21 17:07:56 UTC
Permalink
Post by William Enestvedt
P.S. That A=W/V equation looks familiar. I just wish I could remember
that kind of stuff. :7( Mind you, for some reason, I haven't been able
to get PV=nRT out of my head since tenth grade, though I'm hazy on what
use it is to me now that I don't measure anything in moles or
atmospheres anymore...
That's the ideal gas law.
--
Same geek, same site, new location
Phil Stracchino Landline: 603-429-0220
***@speakeasy.net Mobile: 603-216-7037
Renaissance Man, Unix generalist, Perl hacker, Free Stater
Steve Hatle
2006-09-21 17:16:53 UTC
Permalink
Post by Phil Stracchino
That's the ideal gas law.
I thought that was "He who smelt it, dealt it"

:-)
Lionel Peterson
2006-09-21 17:49:48 UTC
Permalink
Date: 2006/09/21 Thu PM 12:07:56 CDT
Subject: Re: [rescue] Real amp draw of E3000
Post by William Enestvedt
P.S. That A=W/V equation looks familiar. I just wish I could remember
that kind of stuff. :7( Mind you, for some reason, I haven't been able
to get PV=nRT out of my head since tenth grade, though I'm hazy on what
use it is to me now that I don't measure anything in moles or
atmospheres anymore...
That's the ideal gas law.
It's stuck in my head as "PV equals NOTHING but nRT" - then again, I barely passed HS chemistry (my labwork carried the day, my test scores were abysmal - I had some sort of mental block), and later failed to achieve a passing grade in two subsequent runs at College Chemistry, causing me to run screaming from the cold, hard engineering school in Hoboken and retreat into the gentler, comforting bosom of a liberal arts education in central Pennsylvania...

Lionel
William Enestvedt
2006-09-21 18:21:05 UTC
Permalink
Post by Phil Stracchino
That's the ideal gas law.
Right, right. *nervous smile* But why can't I get it out of my head?
Not like it's got a catchy beat or something, it just won't. Go. Away.

Two weeks ago I did some honest-to-gosh algebra, and felt quite
celver -- but this... it's wearing out its welcome, I can tell you.

- Will
--
Will Enestvedt
UNIX System Administrator
Providence, RI
William Enestvedt
2006-09-21 19:58:06 UTC
Permalink
Pat wrote:
--->8---SNIP--->8-_hard math stuff snipped_---SNIP---8<---
Post by Patrick Finnegan
Will, any chance you got any of that? :)
*pulls head out of refrigerator, beer in hand* "I'm sorry...what? Did
yYou want one of these?"

- Will
--
Will Enestvedt
UNIX System Administrator
Providence, RI
William Enestvedt
2006-09-21 19:59:37 UTC
Permalink
... I watched half a 'data centre' loose power twice in 5 hours
because of loading issues (and a *very* big burn out fuse).
On the 2005 Father's Day weekend (which was also my birthday), a rack
of my core production servers did that dance on a set of PDUs. One PDU's
breaker would trip, every server's whole load would fall over to its
other PSU, and the next PDU would die as the first came back up. It
happened for a couple of hours before I got in -- luckily nothing died
permanently. (Good enough servers, those Sun 480s.)

"Racking creep" had ended up with four servers and two RAIDs on two
(three, maybe, I forget) 15A PDUs. I should have seen it coming! *blush*
I ended up putting four 20A PDUs into the rack on four separate supply
circuits.

Anyway, I just realized I had a rack of three-phase gear which still
had both open spaces and an empty, live 15A PDU. So the V40z has a new
home, atop some TEST servers.

Thank you to everyone who stayed tuned. And I am now going to expand
the catalog I made last year of our heat output (when a chiller died in
August -- don't ask...) to include power draw at maximum rated, and try
to get the nominal draw measured by our
former-missle-tech-turned-too-often-bored Wang operator. Maybe he can
teach me some more of this using puppets or something. :7)

- Will
--
Will Enestvedt
UNIX System Administrator
Providence, RI
William Kirkland
2006-09-22 21:33:46 UTC
Permalink
Post by William Enestvedt
P.S. That A=W/V equation looks familiar. I just wish I could remember
that kind of stuff. :7( Mind you, for some reason, I haven't been able
to get PV=nRT out of my head since tenth grade, though I'm hazy on
what
use it is to me now that I don't measure anything in moles or
atmospheres anymore...
... UGLY's electrical references, isbn=0-9623229-4-6

--
***@gmail.com
Bryan Gurney
2006-09-23 00:39:10 UTC
Permalink
On Fri, 22 Sep 2006 17:33:46 -0400, William Kirkland
Post by William Kirkland
Post by William Enestvedt
P.S. That A=W/V equation looks familiar. I just wish I could remember
that kind of stuff. :7( Mind you, for some reason, I haven't been able
to get PV=nRT out of my head since tenth grade, though I'm hazy on
what
use it is to me now that I don't measure anything in moles or
atmospheres anymore...
... UGLY's electrical references, isbn=0-9623229-4-6
Just in case you haven't already had E=IR and P=VI hammered into your head
by four years of college.

Too bad I don't get to use it that much as a sysadmin; after I graduated,
the computer-related electronics field in the area basically said to me,
"You don't have a master's, you don't have any inventions, and you don't
have any filed patents; you are certified as Not Economically Viable, now
go away."

I know; it's who you know, but those who I did know didn't have enough
money.

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