I’ve been researching the load for my servers in our new machine room at the CU Stadium. I use the Dell Rack Advisor to calculate heat and power in BTU and Amps. I like it because it tells you how you should be loading your rack bottom to top and gives you power and heat information down to things like the fan kit. It will also tell you how many power strips you need and part numbers for everything. Unfortunately the Dell Rack Advisor 5.0 does not include current servers like the 1850, 2850 etc. For now I’m guessing on the few 2850s we have and comparing them to a 2650. Maybe we could figure out a ratio to use vs Dells operating at a DOS prompt temperature using the values from the rack advisor. Use google to look up rack advisor 5.0.

Dell’s new version - Dell Product Configuration Calculator - has information about all of the newer servers, as well as new peripheral items and storage equipment. This configurator returns more detailed information than the rack advisor does, however some of the older models such as the PE1550, etc., are not available. Your best bet is to use both if you have various machines in your environment. Go to Dell.com and look up Product Calculator.

Is there any formula to calculate heat dissipation of a server? what is the basic inputs required to calculate BTU of a server. I need to design the AC requirement for my data center and hence this is most important. Note i have server of various model ranging from 4 years old to recent one.

I’ve been told that MAX Watts being used multiplied by 3.413 provides BTU/hr.
Not all machines use the entire Wattage available in their Power Supply, but it doesn’t hurt to over estimate cooling requirements.
Then divide the BTU totals for room by 12000, to get Cooling Tonnage.

[tags]Thermal Load, BTU/hour[/tags]

I dug up this information on the PowerEdge 1750

PowerEdge 1750:
Power supply:
Wattage: 320 W (AC)
Voltage:100–240 VAC, 50/60 Hz, 3.9–2.0 A
Heat dissipation: 1026 BTU/hr maximum per power supply <—- notice the “per” power supply
Maximum inrush current
Under typical line conditions and over the entire system ambient operating range, the inrush current may reach 25 A per power supply for 10 ms or less.
System battery:3.0-V lithium ion coin cell

and then similarly for the Power Edge 1850, because I needed to compute heat dissipation for my company’s datacenter.

Power Edge 1850:
Power
AC power supply (per power supply)
Wattage 550 W
Voltage 84 -264 VAC, autoranging, 47-63 Hz, 7.6 A
Heat dissipation 2130 BTU/hr (theoretical maximum)

If you would notice, the PE1850 theoretical maximum is over 2000 BTUs because these blades have two redundant power supplies. Whether they are both in use or not at the same time is the question I am unsure of, so I just gave the datacenter 2100 max BTUs as the heat dissipation. I really wish Dell would be more clear about this kind of stuff.

The APC web site contains a calculator for how many UPS battery packs are needed. It shows for

Dell PowerEdge 2950

Total Power (Watts): 374
# of Power Cords: 2
Total Power (VA): 534
Quantity: 1
Total Thermal (BTU/hr): 1277

http://www.apc.com/tools/ups_selector/index.cfm?action=select_device_mfg&appUsed=advanced&device_type=server&mfg_filter=ALL&app_type=server&showManufacturer=1&showChassis=1&showMonitor=1&showProcessor=1&showNumPwrSupplies&showNumProcessors=1&showVoltages=1&showIntDrives=1&showExtDrives=1&showHardDrives=1&showNumPCISlots=1&showNumPwrCords=1&showPlugType=1&showQty=1&showPeripherals=1&config_id=00000000-0000-0000-0000-000000000000&cgi_referer_check=&return_query_string=

Hay guys i am just curious in order to minimize heat, would it be better to load server racks from top bottom or bottom up? does this make much of a difference if i have 5x 2RU serverrs in a large 45 RU rack?

Thank you

Check out http://www.dell.com/calc for BTU calculations for Dell servers.

I took the post from Gavin and did the following calculation:

Total Power: 374 Watts
Total Thermal: 1277 BTU/hr.

Since there is 3413 BTU/hr in 1kW, we have:

.374 kW * 3413 (BTU/hr.)/1kW = 1276.462 BTU/hr.

Which is almost exactly what the specs say for total thermal dissipation.

Meaning, 99% of the total power consumed is dissipated as heat. This is in line with what you read about servers being only 1% efficient.

Therefore, you can take the total power your servers use, in watts, and just divide by 3517.2 to get total tons refrigeration needed. 1 ton of refrigeration is equivalent to 3517.2 watts.

I _think_ I got this down right. Someone correct me if I’m wrong.

The tonnes you discuss Garry are Sensible Tonnes. Heat load is divided into Sensible and Latent.

Sensible+Latent=Total

Your AC might have a sensible heat ratio of say 0.73 to 0.95. Let’s say your AC unit has a 0.85 sensible heat ratio.

You need 1 ton of sensible heat cooling, therefore you need a unit sized at 1 / 0.85 = 1.18 Cooling Tons Nominal. This becomes crucial when you have say 50 tons of load in a computer room.