WSJ: Big tech firms seeking power
Since power consumption was a topic at the last NANOG meeting. subscription required, or buy a copy of the Wall Street Journal from a newstand http://online.wsj.com/article/SB115016534015978590.html Surge in Internet Use, Energy Costs Has Big Tech Firms Seeking Power By KEVIN J. DELANEY and REBECCA SMITH Wall Street Journal June 13, 2006; Page A1 With both Internet services and power costs soaring, big technology companies are scouring the nation to secure enough of the cheap electricity that is vital to their growth. The search is being led by companies including Microsoft Corp., Yahoo Inc. and IAC/InterActiveCorp. Big Internet firms have been adding thousands of computer servers to data centers to handle heavy customer use of their services, including ambitious new offerings such as online video. [...]
On 6/14/06, Sean Donelan <sean@donelan.com> wrote:
Since power consumption was a topic at the last NANOG meeting.
subscription required, or buy a copy of the Wall Street Journal from a newstand
http://online.wsj.com/article/SB115016534015978590.html Surge in Internet Use, Energy Costs Has Big Tech Firms Seeking Power By KEVIN J. DELANEY and REBECCA SMITH Wall Street Journal June 13, 2006; Page A1
With both Internet services and power costs soaring, big technology companies are scouring the nation to secure enough of the cheap electricity that is vital to their growth.
The search is being led by companies including Microsoft Corp., Yahoo Inc. and IAC/InterActiveCorp. Big Internet firms have been adding thousands of computer servers to data centers to handle heavy customer use of their services, including ambitious new offerings such as online video. [...]
And, just to be fair, Google gets their own bit of news on the power front: http://www.iht.com/articles/2006/06/13/business/search.php I wonder just how much power it takes to cool 450,000 servers. Matt
I wonder just how much power it takes to cool 450,000 servers.
I've heard mumbles that the per kWh rates from Bonneville in the locations along the Columbia are in the sub-4¢ range. Grant county is seeing a huge fiber building boom as a result. It will be more wired up than King county soon. Woody was here last night and remarked (feel free to correct me if I misquote you Bill) that it was funny that nowadays "network geeks were more interested in kilowatts than kilobits" --chuck (in Seattle)
450,000 * 100 WT (power itself) Cooling - I donot know, but I should estimate it as extra 70% of consumed power. So, 450,000 * 0.2KWT = 90,000KWT. ----- Original Message ----- From: "chuck goolsbee" <chucklist@forest.net> To: <nanog@merit.edu> Sent: Friday, June 16, 2006 10:47 AM Subject: Re: WSJ: Big tech firms seeking power
I wonder just how much power it takes to cool 450,000 servers.
I've heard mumbles that the per kWh rates from Bonneville in the locations along the Columbia are in the sub-4¢ range. Grant county is seeing a huge fiber building boom as a result. It will be more wired up than King county soon. Woody was here last night and remarked (feel free to correct me if I misquote you Bill) that it was funny that nowadays "network geeks were more interested in kilowatts than kilobits" --chuck (in Seattle)
This article talks about power and costs: http://www.post-gazette.com/pg/06164/697875-96.stm Interesting the power today is being used for cold storage and Aluminum plants because it is so cheap. Christian -----Original Message----- From: owner-nanog@merit.edu [mailto:owner-nanog@merit.edu] On Behalf Of chuck goolsbee Sent: Friday, June 16, 2006 10:48 AM To: nanog@merit.edu Subject: Re: WSJ: Big tech firms seeking power
I wonder just how much power it takes to cool 450,000 servers.
I've heard mumbles that the per kWh rates from Bonneville in the locations along the Columbia are in the sub-4¢ range. Grant county is seeing a huge fiber building boom as a result. It will be more wired up than King county soon. Woody was here last night and remarked (feel free to correct me if I misquote you Bill) that it was funny that nowadays "network geeks were more interested in kilowatts than kilobits" --chuck (in Seattle)
I wonder just how much power it takes to cool 450,000 servers.
450,000 servers * 100 Watts/Server = 45,000,000 watts / 3.413 watts/ BTU = 13.1 Million BTU / 12000 BTU/Ton = 1100 Tons of cooling A 30 Ton Liebert system runs about 80 amps @ 480 volts or 38400 watts, you'll need at least 40 or them to cool 1100 tons which is 1536 Kw * 24 hours * 7 days * 4.3 weeks = 1,110,000 KwH/month * $0.10/ KwH = $111,000 /month in cooling. I think my math is right on this... -- Matthew S. Crocker Vice President Crocker Communications, Inc. Internet Division PO BOX 710 Greenfield, MA 01302-0710 http://www.crocker.com
On Fri, 16 Jun 2006, Matthew Crocker wrote:
I wonder just how much power it takes to cool 450,000 servers.
450,000 servers * 100 Watts/Server = 45,000,000 watts / 3.413 watts/BTU = 13.1 Million BTU / 12000 BTU/Ton = 1100 Tons of cooling
Error: you MULTIPLY 3.413 to go from watts to BTU, not divide. It's be more like 154,000,000 BTU, /12000 or 12,798 tons. Also at 100 watts, you are assuming Celerons with single hard drives. We see more like 120 to 240 depending on config. 100 would be low.
A 30 Ton Liebert system runs about 80 amps @ 480 volts or 38400 watts, you'll need at least 40 or them to cool 1100 tons which is 1536 Kw * 24 hours * 7 days * 4.3 weeks = 1,110,000 KwH/month * $0.10/KwH = $111,000 /month in cooling.
80 amps @ 480 is 80 * 480 * 1.73, or 66 kw. However, they don't draw that much. A 30 ton unit, worst case (115 degrees outside across the condensor) will be about 50 kw, assuming you do not have humidification or reheats turned on. Second issue: you are assuming 100% cooling efficiency, or, in other words, that you'd have perfect airflow, perfect air return, etc. Never happens, especially when you have customers who are idiots. Third issue: you are assuming there is no heat loss or gain in the structure of the building. This could be very significant. Let's assume it's not. It's likely in an environment like this, you'd have more like 14000 tons. 14000 / 30 = 466 units, @ 50 kw/unit, 23,300,000 watts, / 1000 * 24 * 30.4375 (avg days in a month) = 17,020,000 kw-hrs, @ $0.12 (more likely with todays fuel prices unless you are in Kentucky) $2,042,400/month. Also, don't forget the original 450,000 servers at 100 watts (45 mw) would be $3,944,700/month in power. Also, 450,000 1U servers at 40/rack would be 11,250 racks, which at 10 sq-ft a rack would be 112,000 sq-ft of datacenter floor space (triple or, more likely, quadruple that for space for HVAC, generators, switchgear, UPSs, etc). That'd be 500,000 sq-ft at minimum. Total is $5,987,000/mon, but you haven't ROIed the millions in electrical gear (think big: this is about 68 megawatts; $250k/each for a 2 mw generator (you'd need 40, $10 mm), $50k/each for a 500 kva UPS (you'd need 80 $4mm), millions in panels, breakers, piping, copper wire (700% increase in copper pricing in the last 24 months, people), etc. Oh, and 466 liebert 30 ton HVAC's, probably $25 to $40k/ea installed ($11 million). Oh, and no one has installed it yet, and you haven't paid rent on the facility that will take 2 years to build with probably 100's of workers saleries. Take $6mm/month, divide by 450,000 servers, $13.33/month/server. Oh, and 68 Megawatts over 112k ft of floor space is 607 watts/ft. Thats about 6 times what most centers built in the last couple years are built at. But wait, there is more. Just a point of comparison -- Oyster Creek Nuclear Power generation plant, located here on the Jersey Shore, produces 636 megawatts. You'd take one-tenth of that capacity -- in a bulding that would sit on a 10 or 20 acre chunk of land. I put this into the 'unlikely' category. The substation alone to handle stepping 68 mwatts from transmission to 480v would be probably 4 acres. And, 68 megawatts of power at 480 volts 81,888 amps. A typicall 200,000 sq-ft multi-tenant office building has 1600 amps of service; this would be the equivalent of 50 buildings. Having fun yet? A 30 ton liebert takes about 30 sq-ft of floor space; 466 of them would be 13,980 sq-ft. If you use a drycooler system, they are about 100 sq-ft, and youd need 233 of them (60 ton DDNT940's), 23,300 sq-ft of roof space. Each of those weighs 2,640 pounds, for a total of 615,000 pounds, or 308 tons (of weight, not HVAC capacity). I won't even spend the CPU cycles figuring out how many gallons of glycol this would bem but probably a good guess would be about 50,000 gallons. That'd be about a quarter-million dollars in glycol. I'm tired now, time to climb back in my hole. In other words, don't get me started on the datacenter density issue. -- Alex Rubenstein, AR97, K2AHR, alex@nac.net, latency, Al Reuben Net Access Corporation, 800-NET-ME-36, http://www.nac.net
--On June 16, 2006 5:24:27 PM -0400 Alex Rubenstein <alex@nac.net> wrote:
But wait, there is more. Just a point of comparison -- Oyster Creek Nuclear Power generation plant, located here on the Jersey Shore, produces 636 megawatts. You'd take one-tenth of that capacity -- in a bulding that would sit on a 10 or 20 acre chunk of land. I put this into the 'unlikely' category. The substation alone to handle stepping 68 mwatts from transmission to 480v would be probably 4 acres. And, 68 megawatts of power at 480 volts 81,888 amps. A typicall 200,000 sq-ft multi-tenant office building has 1600 amps of service; this would be the equivalent of 50 buildings.
Having fun yet?
I happen to know that a very large power line project was just finished in that area :) (I have family that works for the company that did the job). It's a huge amount of power that's for sure. I'm not sure what the exact route was, nor the endpoint right now, but when I did ask him at the time it didn't make sense....Now it might. I'll talk to him again.
Seen few data centers: - biggerst cages are about 500 servers, may be you can pack 1,000 servers; - ok, how many cagses in the medium size building? I''d say, 100 - 200 (may be less). So, 1 building can handle 50,000 - 100,000 servers. A very big building, I guess, can handle 450,000. But what for? You can put it al together, but how you deliver input data and ship output data from 450,000 servers? ----- Original Message ----- From: "Michael Loftis" <mloftis@wgops.com> To: "Alex Rubenstein" <alex@nac.net>; <nanog@merit.edu> Sent: Friday, June 16, 2006 3:32 PM Subject: Re: WSJ: Big tech firms seeking power
--On June 16, 2006 5:24:27 PM -0400 Alex Rubenstein <alex@nac.net> wrote:
But wait, there is more. Just a point of comparison -- Oyster Creek Nuclear Power generation plant, located here on the Jersey Shore, produces 636 megawatts. You'd take one-tenth of that capacity -- in a bulding that would sit on a 10 or 20 acre chunk of land. I put this into the 'unlikely' category. The substation alone to handle stepping 68 mwatts from transmission to 480v would be probably 4 acres. And, 68 megawatts of power at 480 volts 81,888 amps. A typicall 200,000 sq-ft multi-tenant office building has 1600 amps of service; this would be the equivalent of 50 buildings.
Having fun yet?
I happen to know that a very large power line project was just finished in that area :) (I have family that works for the company that did the job). It's a huge amount of power that's for sure. I'm not sure what the exact route was, nor the endpoint right now, but when I did ask him at the time it didn't make sense....Now it might. I'll talk to him again.
On 6/16/2006 at 2:24 PM, Alex Rubenstein <alex@nac.net> wrote: On Fri, 16 Jun 2006, Matthew Crocker wrote:
I wonder just how much power it takes to cool 450,000 servers.
450,000 servers * 100 Watts/Server = 45,000,000 watts / 3.413
watts/BTU =
13.1 Million BTU / 12000 BTU/Ton = 1100 Tons of cooling
Error: you MULTIPLY 3.413 to go from watts to BTU, not divide. It's be more like 154,000,000 BTU, /12000 or 12,798 tons.
Well, the bigger problem here is that a watt is a measure of power (engergy/time) and a BTU is a unit of energy. There is no dimensionless conversion factor between the two. -- Crist J. Clark crist.clark@globalstar.com Globalstar Communications (408) 933-4387 B¼information contained in this e-mail message is confidential, intended only for the use of the individual or entity named above. If the reader of this e-mail is not the intended recipient, or the employee or agent responsible to deliver it to the intended recipient, you are hereby notified that any review, dissemination, distribution or copying of this communication is strictly prohibited. If you have received this e-mail in error, please contact postmaster@globalstar.com
On Fri, 16 Jun 2006, Crist Clark wrote:
Error: you MULTIPLY 3.413 to go from watts to BTU, not divide. It's be more like 154,000,000 BTU, /12000 or 12,798 tons.
Well, the bigger problem here is that a watt is a measure of power (engergy/time) and a BTU is a unit of energy. There is no dimensionless conversion factor between the two.
Huh? A Watt has no time constant. A watt is an amount of energy consumed at a moment (ie, a 60 watt light bulb), not an amount of energy over time (like a watt-hour; for instance, a 60 watt light bulb uses 60 watt-hours of power every hour, or 1.44 kwatt-hrs per day). There is a direct correlation between watts and btu's, and that is: watts * 3.413 = btu -- Alex Rubenstein, AR97, K2AHR, alex@nac.net, latency, Al Reuben Net Access Corporation, 800-NET-ME-36, http://www.nac.net
Alex Rubenstein wrote:
Huh?
A Watt has no time constant. A watt is an amount of energy consumed at a moment (ie, a 60 watt light bulb), not an amount of energy over time (like a watt-hour; for instance, a 60 watt light bulb uses 60 watt-hours of power every hour, or 1.44 kwatt-hrs per day).
There is a direct correlation between watts and btu's, and that is:
watts * 3.413 = btu
You're confusing Watts and joules. One Watt is one joule of energy per second.
On Fri, 16 Jun 2006, Alex Rubenstein wrote:
more like 154,000,000 BTU, /12000 or 12,798 tons.
Well, the bigger problem here is that a watt is a measure of power (engergy/time) and a BTU is a unit of energy. There is no dimensionless conversion factor between the two.
Huh?
A Watt has no time constant. A watt is an amount of energy consumed at a moment (ie, a 60 watt light bulb), not an amount of energy over time (like a watt-hour; for instance, a 60 watt light bulb uses 60 watt-hours of power every hour, or 1.44 kwatt-hrs per day).
Since you like Wikipedia so much, why don't you look it up: http://en.wikipedia.org/wiki/Watt Watt is not amount of power but amount of power produced during time, i.e. its speed of energy consumption. However kwatt-hour (I've never heard of watt-hour, but I suppose that maybe used too..) is actually amount of energy consumed - more precisely X kwr its how much energy device would consume if it were consuming energy at exactly the same speed of X kw for entire hour. -- William Leibzon Elan Networks william@elan.net
On Fri, Jun 16, 2006 at 07:09:37PM -0700, william(at)elan.net wrote:
Watt is not amount of power but amount of power produced during time, i.e. its speed of energy consumption.
Actually, that's the definition of power. (Energy/time) http://en.wikipedia.org/wiki/Electric_power A kilowatt-hour is equivalent to 3600000 joules. --Adam
Once upon a time, Alex Rubenstein <alex@nac.net> said:
There is a direct correlation between watts and btu's, and that is:
watts * 3.413 = btu
No, that's wrong. $ units 2438 units, 71 prefixes, 32 nonlinear units You have: watt You want: btu conformability error 1 kg m^2 / s^3 1055.0559 kg m^2 / s^2 You have: watt hour You want: btu * 3.4121416 / 0.29307107 -- Chris Adams <cmadams@hiwaay.net> Systems and Network Administrator - HiWAAY Internet Services I don't speak for anybody but myself - that's enough trouble.
watts * 3.413 = btu
No, that's wrong. ..... You have: watt hour You want: btu * 3.4121416 / 0.29307107
Rant: After I get the low-bid subcontract to manage the place; I'm going to set aside a special section Down There just for the HVAC folks who insist on perpetuating that most medieval of units... the BTU. It belongs in the pile of toxic waste that now holds farthings, stone, furlongs, and slugs. Some day, I'll be able to look at EPA Yellow Tags on water heaters and AC's without grinding my teeth.. /Rant -- A host is a host from coast to coast.................wb8foz@nrk.com & no one will talk to a host that's close........[v].(301) 56-LINUX Unless the host (that isn't close).........................pob 1433 is busy, hung or dead....................................20915-1433
On Fri, 16 Jun 2006, David Lesher wrote:
who insist on perpetuating that most medieval of units... the BTU.
Well, if you do away with that you can continue with the "mile" as well, then lose the pounds and yards and gallons while you're at it. On the other hand, I have a question I was pondering at the nanog power session (which was a really good one). What is the amount of energy coming out of a server as heat as opposed to what you put in as electricity? My guess would be pretty close to 100%, but is it really so? And I've also been told that you need approx 1/3 of the energy taken out thru cooling to cool it? So that would mean that to sustain a 100W server you really need approx 130-140W of power when cooling is included in the equation. Is this a correct assumption? In one of our data centers we use community cooling, we get 4 C (I think it was approx 4 C) degree water and we're required to heat it at least by 8 C before we return it, this is then used in the community power plant to produce hot community water, and this process I've been told is quite effective. Any thoughts on this? Guess it doesn't work in the boondocks though... There were also plans to use heat converters to have the cooling water from nuclear power plants heat community hot water, but politicians chickened out. Now we just spew that cooling water straight out into the ocean. I guess none of this makes sense in the southern part of the US, but further up north where houses actually need heating and not cooling most of the year, are things like this done? -- Mikael Abrahamsson email: swmike@swm.pp.se
What is the amount of energy coming out of a server as heat as opposed to what you put in as electricity? My guess would be pretty close to 100%, but is it really so? And I've also been told that you need approx 1/3 of the energy taken out thru cooling to cool it? So that would mean that to sustain a 100W server you really need approx 130-140W of power when cooling is included in the equation. Is this a correct assumption?
Based upon my real-world experience, and talking to a few folks, it's very close to 100%. Most assume 100% for the practice of calculating cooling. However, for those who are very scientific, they try to tell you that some of the power is going into movement of hard drive heads, etc., which creates force on your racks, etc. A true, but irrelevant discussion, really, because it's likely an immeasurable amount. One could do the excercise of putting a computer in a well insulated box and measuring power in vs. rate of rise of temperature. Volunteers? :) -- Alex Rubenstein, AR97, K2AHR, alex@nac.net, latency, Al Reuben Net Access Corporation, 800-NET-ME-36, http://www.nac.net
Speaking on Deep Background, the Press Secretary whispered:
who insist on perpetuating that most medieval of units... the BTU.
Well, if you do away with that you can continue with the "mile" as well, then lose the pounds and yards and gallons while you're at it.
Great! Let's get started...
What is the amount of energy coming out of a server as heat as opposed to what you put in as electricity? My guess would be pretty close to 100%,
Actually, it's closer to 100.00%. Most is heat directly, but some very small amount is mechanical work on the HD's, etc... and that is then room heat as the drives radiate heat to the room. The LED's emit photons that then heat the room, and so forth. The only energy that 'escapes' the building would likely be outgoing copper & glass data connections....but wait, there's incoming of THOSE too. [But... if a server farm, there's more bits out than in...]
In one of our data centers we use community cooling, we get 4 C (I think it was approx 4 C) degree water and we're required to heat it at least by 8 C before we return it, this is then used in the community power plant to produce hot community water, and this process I've been told is quite effective. Any thoughts on this? Guess it doesn't work in the boondocks though...
I guess none of this makes sense in the southern part of the US, but further up north where houses actually need heating and not cooling most of the year, are things like this done?
Almost never. In the immediate focus of the US it's cheaper to import foreign oil/mine & burn coal than to invest capital to do something more efficiently. [Hmmm, I wonder what the current power price is in the Niagara River Valley? Their cheap power was why so many steel mills/aluminum smelters/etc located there eons ago. Plus, there's copious H2O cooling and I have to think there are massive buildings available in the area just for paying the back taxes...] [This has drifted way OT and I'm out of here...] -- A host is a host from coast to coast.................wb8foz@nrk.com & no one will talk to a host that's close........[v].(301) 56-LINUX Unless the host (that isn't close).........................pob 1433 is busy, hung or dead....................................20915-1433
Mecahnical work converts to heat in the very end. Not _mostly 100%_ but _absolutely 100%_. Except if it is cell station which inducts energy into the radio wawes, and minus some light coming out of the building (which removes energy as well). ----- Original Message ----- From: "David Lesher" <wb8foz@nrk.com> To: "nanog list" <nanog@merit.edu> Sent: Saturday, June 17, 2006 4:08 AM Subject: Re: WSJ: Big tech firms seeking power
Speaking on Deep Background, the Press Secretary whispered:
who insist on perpetuating that most medieval of units... the BTU.
Well, if you do away with that you can continue with the "mile" as well, then lose the pounds and yards and gallons while you're at it.
Great! Let's get started...
What is the amount of energy coming out of a server as heat as opposed
to
what you put in as electricity? My guess would be pretty close to 100%,
Actually, it's closer to 100.00%. Most is heat directly, but some very small amount is mechanical work on the HD's, etc... and that is then room heat as the drives radiate heat to the room. The LED's emit photons that then heat the room, and so forth.
The only energy that 'escapes' the building would likely be outgoing copper & glass data connections....but wait, there's incoming of THOSE too. [But... if a server farm, there's more bits out than in...]
In one of our data centers we use community cooling, we get 4 C (I think it was approx 4 C) degree water and we're required to heat it at least by 8 C before we return it, this is then used in the community power plant to produce hot community water, and this process I've been told is quite effective. Any thoughts on this? Guess it doesn't work in the boondocks though...
I guess none of this makes sense in the southern part of the US, but further up north where houses actually need heating and not cooling most of the year, are things like this done?
Almost never. In the immediate focus of the US it's cheaper to import foreign oil/mine & burn coal than to invest capital to do something more efficiently.
[Hmmm, I wonder what the current power price is in the Niagara River Valley? Their cheap power was why so many steel mills/aluminum smelters/etc located there eons ago. Plus, there's copious H2O cooling and I have to think there are massive buildings available in the area just for paying the back taxes...]
[This has drifted way OT and I'm out of here...]
-- A host is a host from coast to coast.................wb8foz@nrk.com & no one will talk to a host that's close........[v].(301) 56-LINUX Unless the host (that isn't close).........................pob 1433 is busy, hung or dead....................................20915-1433
No, that's wrong.
$ units 2438 units, 71 prefixes, 32 nonlinear units
You have: watt You want: btu conformability error 1 kg m^2 / s^3 1055.0559 kg m^2 / s^2 You have: watt hour You want: btu * 3.4121416 / 0.29307107
Agreed, my math should have said "btu/hr", which is what any HVAC system is rated in -- how many btus in an hour it can remove. I apologize for the horrendous error, but all of the math stands. Just sed s/btu/btu\/hr/g (also, you can do from watt to btu/hr with the same 3.413 multiplier) -- Alex Rubenstein, AR97, K2AHR, alex@nac.net, latency, Al Reuben Net Access Corporation, 800-NET-ME-36, http://www.nac.net
On Fri, 16 Jun 2006, Chris Adams wrote:
Once upon a time, Alex Rubenstein <alex@nac.net> said:
There is a direct correlation between watts and btu's, and that is:
watts * 3.413 = btu
No, that's wrong.
Oh lord. a Watt is equal to one joule of energy per second. Period. a watt/hour is equal to, oh, lets see, 3600 joules consumed in that hour. 1 joule is oh, 0.00094781712 btu according to one chart, .00094845 on another. the math is really straight forward here guys... --- david raistrick http://www.netmeister.org/news/learn2quote.html drais@atlasta.net http://www.expita.com/nomime.html
On Fri, 16 Jun 2006, Crist Clark wrote:
Error: you MULTIPLY 3.413 to go from watts to BTU, not divide. It's be more like 154,000,000 BTU, /12000 or 12,798 tons.
Well, the bigger problem here is that a watt is a measure of power (engergy/time) and a BTU is a unit of energy. There is no dimensionless conversion factor between the two.
Alright, I am sorry I missed that. It should read:
Error: you MULTIPLY 3.413 to go from watts to BTU/hr, not divide. It's be more like 154,000,000 BTU/hr, /12000 or 12,798 tons.
Sorry! Sheesh. -- Alex Rubenstein, AR97, K2AHR, alex@nac.net, latency, Al Reuben Net Access Corporation, 800-NET-ME-36, http://www.nac.net
I used very raw estimation (which is not well correct but dont make too much of errors) - to remive 1 KW out of building, yiou spend extra 1 KW. But anyway, 450,000 servers have a great power consumption - you can use river or a lake to cool them, but you still need 45,000 KW of power to make them work. So, to say 60,000KW - 100,000KW will nopt be a big mistake (total power consumption). Of course, if you build data center inside the the power plant dam, then you have both, enough cooling and enough power. -:) ----- Original Message ----- From: "Matthew Crocker" <matthew@crocker.com> To: <nanog@merit.edu> Sent: Friday, June 16, 2006 1:16 PM Subject: Re: WSJ: Big tech firms seeking power
I wonder just how much power it takes to cool 450,000 servers.
450,000 servers * 100 Watts/Server = 45,000,000 watts / 3.413 watts/ BTU = 13.1 Million BTU / 12000 BTU/Ton = 1100 Tons of cooling
A 30 Ton Liebert system runs about 80 amps @ 480 volts or 38400 watts, you'll need at least 40 or them to cool 1100 tons which is 1536 Kw * 24 hours * 7 days * 4.3 weeks = 1,110,000 KwH/month * $0.10/ KwH = $111,000 /month in cooling.
I think my math is right on this...
-- Matthew S. Crocker Vice President Crocker Communications, Inc. Internet Division PO BOX 710 Greenfield, MA 01302-0710 http://www.crocker.com
participants (16)
-
Adam McKenna -
Alex Rubenstein -
Alexei Roudnev -
Chris Adams -
Christian Nielsen -
chuck goolsbee -
Crist Clark -
David Lesher -
david raistrick -
Matthew Crocker -
Matthew Petach -
Michael Loftis -
Mikael Abrahamsson -
Nicholas Suan -
Sean Donelan -
william(at)elan.net