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Data Center Toolbox for Power and Cooling

Posted by John Musilli on Sep 27, 2007 6:20:00 PM

Things you need to operate a successful Data Center infrastructure.

 

This is a first in a series of Toolbox topics others include

"Watts per Sq.Ft.of What"

"Use of a Hand Held IR (Infra Red) Gun for a Data Center Health Check"

"Generic Data Center Racking, Cost and Space Benifits"

"Data Center Layer One and Structured Cabling Designs, Without Costly Patch Panel Installations"

 

As a data center operations manager you are reasonable for the stability of the physical infrastructure of your environment. Often this requires support from maintenance and or engineering staff to provide you with capacity and room loading calculations. In order for you to do your job efficiently and not be reliant on others you need a few tools to Help You Help Yourself the first in a series is:

 

*Data Center Math

*Power and Thermal Measurement

 

Watts (w) = Volts (V) x Amps (A)

 

Voltage x Amps +=+ KW (Kilo Watts) (This is Electrical Heat)

1000 w

 

British Thermal Unit (BTU) (measure of heat)

 

One Watt of Power requires 3.432 BTU's to cool

 

12,000 BTU's = One Ton of Cooling

 

Example;

120 Volts x 160 Amps = 19,200 Watts = 19.2 kW

 

 

19,200w x 3.432 BTU = 65,894 BTU = 5.5 Tons of Cooling Required

One Ton of Cooling = 12,000 BTU

 

 

Power Basics:

Reduce all loads to Watts as the common measurement including cooling. If you use Watts as the common unit you do not need Amps or Voltage when determining capacities.

 

 

 

Power Rough Rules of Thumb

• One average rack of 2u to 8u servers, (40u's total) use~ 5000watts

• One disc type storage bay(24inches)is ~5000watts

• One network equipment rack ~ 30 to 40u's of switches requires 5000w to 6000w

• The average server landing power requirement with redundant network and redundant disc storage is 400watts per server

• The average server landing power requirement with single network switch and single storage connectivity is 300watts per server

• The average "One U" server rack with 40 servers per rack ranges between 7500w to 9000w depending on utilization

• One blade center is 3600w to 4000w

 

Cooling Rough Rule of Thumb

One blade center @3600watts requires 1 ton of cooling.

• One rack of 2u through 8u servers, (40u's total) required 1 1/2 (one and one half) tons of cooling

• The industry standard rack doors can restrict up to 40% of the air flow

• If using "relative humidity set points" set @ 50% plus or minus 20% this will reduce alarms and operating cost

• Available supply air temperature at the server intake can be as high as 80 degrees Fahrenheit without issues

 

 

If this Information is useful please comment

 

 

*Disclaimer

The opinions, suggestions, management practices, room capacities, equipment placement, infrastructure capacity, power and cooling ratios are strictly the opinion and observations of the author and presenter.

The statements, conclusions, opinions, and practices shown or discussed do not in any way represent the endorsement or approval for use by Intel Corporation.

Use of any design practices or equipment discussed or identified in this presentation is at the risk of the user and should be reviewed by your own engineering staff or consultants prior to use.

 

 



Add a comment Leave a comment on this blog post.
Sep 28, 2007 11:46 AM Guest Brad  says:

Very informative. Thanks

Sep 29, 2007 12:07 PM Guest Israel  says:

Nice post, wished i found this information last time I had to build out a DC.

Nov 4, 2007 10:37 PM Guest Yigal  says:

John,

Thank you for this informative post. My experience, however, is that using standard rules of thumb is what gets people in trouble when they use high density computing such as 4-5 blade centers in a rack. Adding 10 tons to the room when adding 10 bladeCenters in 2 cabinets will not be effective. So I would add a cautious note that these rules may not apply for high density. What do you think?

Nov 8, 2007 1:21 AM Guest John Musilli  says in response to Yigal:

Thank you for pointing this out. You are absolutely correct. When you land that concentration of heat in your example 14kw to 18kw (3600w per blade center) you have to have a significant amount of cold air requirements 1400 to 1700 CFM (Cubic Feet per Minuet) delivered to the racks. Even if you had high flow grates in the rack cooling zone you will most likely starve adjacent areas causing isolated “Hot Spots”

 

To all readers Blades and full racks of 1u servers can be handled successfully even in a data center designed at 50w/sf but you will have to use the Rule of Thumb and your head to properly distribute the heat loads.

JM

Nov 12, 2007 10:21 AM Michael Ryan Michael Ryan    says:

Good discussions to real problems. Thanks........

Feb 17, 2008 10:39 PM Guest Listener  says:

Very useful article, thank you

______________

 

Aug 30, 2008 1:26 AM Guest Sandeep  says:

John, I found this article very informative .. Thanks!

 

Have you any experience on using tubes blowing cool air to solve hot spots. The idea is to get cooling to specific areas on a need basis. Do you think this will work. What could be the challenges.

 

Any IR gun you can recommend, is it the same as thermal IR camera or FLIR.

Sep 1, 2008 12:56 PM John Musilli John Musilli    says in response to Sandeep:

Thank you for your comments and questions

Others please provide your comments or observations

 

As for tubes blowing cool air to solve hot spots it is not clear if you are addressing an isolated hotspot in a room, or are you suggesting this as a thermal air distribution design

 

The blowing cool air to resolve “Hot Spots” is a fairly common data center operations solution using spot cooling units ranging in size from one ton (3500 watt) to three ton (10,000watt) portable cooling systems with poor results. The issue is you need to remove the heat generated by the cooling unit which is still in very close proximity (3ft to 5ft) to the equipment you are trying to cool, this physical closeness causing Short Cycling or heat dumping into another line of equipment in the area.

 

A better solution is to use a large fan (> 10,000CFM) to move and or mix the air in front of the equipment from another area that is cooler then the “Hot Spot” this blended air source will mix with the rest of the room air and distribute the heat more uniformity

 

As a thermal air distribution design this has merit with some caveats. Any time you deliver a stream of air directly to the supply air side of the equipment you need to be aware of the velocity and or static pressure. You can alter the intended design of the OEM for air distribution within the equipment.

 

In short I have considered rack and supply air designs that use supply air tubes incorporated into the server racks that deliver cool air directly to the equipment supply intakes through an articulated duct. This configuration is used on some airliners to keep food carts cold when stored in the galley bays.

 

The IR (Infra Red) gun is similar to the camera except you have a very small target area from a square inch to a square foot. The gun will not catch and store an image.

The specifications for a suitable IR gun include a visible laser targeting feature; record the high low temp range, accuracy plus or minus 2 degrees F. the expected price range is $400 to $600.

Mar 20, 2009 12:13 PM Guest hajoo  says:

This is not a math, this is a mish mash....


Watts (w) = Volts (V) x Amps (A)                                              - right

 

Voltage x Amps

------------------------ = KW (Kilo Watts) (This is Electrical Heat)         - wrong

1000 w

 

right:

 

 

Voltage x Amps

------------------------ = KW (Kilo Watts) (This is Electrical Heat)

1000



One Watt of Power requires 3.432 BTU's to cool                          - wrong, Watt is power unit, BTU is unit for
                                                                                                  amount of heat

1 WattHour (no Watt) equals 3.4121 BTUs of cool capacity          - i think that is right.