Does anyone have rough figures on the raio of kw/h used to heat a datacenter, versus that used to cool?
An associate's datacenter shows about 130kva in total load across his two UPSes, but during generator load-tests, he sees 225-250kw load on his gen-set. That gen-set load includes twenty 10-ton compressorized Lieberts, two cooling towers, four pumps. He seems to be using 2/3 as much power to cool as to heat!
Is 100+kw/h a reasonable usage to operate that cooling plant? Hopefully, it would be more efficient with 300kva+ of heat load.
Studies of 22 Data Centers by Lawrence Berkley labs shows that the most Efficient Data Centers use ~25% of their total energy for cooling and the worst were as high as 60%. The average was 37%. Our findings are on the high end of this scale.
Many factors determine how efficient your cooling infrastructure is. Good Airflow, DX vs Chilled Water, Chiller KW/Ton, VFD drives etc. etc
If the critical load is showing 130kva, then assuming a rough power factor of .8, the computer equipment should create a total heat load of 104 kW. (converting volt-amps to watts of load can be complicated and dependent upon type of power and UPS)
On the cooling side, and first of all, ten 20 ton CRACs is a bunch of cooling (703kW). Cooling is never 100% efficient, but way too much cooling for the heat load. Thus, I'll assume the CRACs are variable (or some off) and 100% efficient, so the net of the CRACs must create 104kW of cooling.
A typical CRAC can have a coefficient of performance (COP) of 3. Thus it can deliver 3 watts of cooling for every watt of input power. To get 104 kilowatts of cooling, the CRAC system requires roughly 35kW of input power.
Cooling towers are pretty efficient and a COP of 7 is reasonable. The heat of the computer gear and the CRACs gets dumped to the cooling tower (and other stuff like UPS and PDU). Again assuming 100% efficiency, the load on the cooling tower should be at least 139kW. That system would require roughly 20kW of input power.
So in a perfect world, to cool the 130kVA (104kW) critical load -- the cooling system would likely require around 55kW of input electrical power. The cooling power input would be 53% of the critical load. At the facility level, the cooling power would be (very rough because I am excluding electrical losses) 35% of the total facility power to the data center.
This is rough, and ideal, but should be in the ball park. Put CRACs a long way from IT gear, go over 150 watts/sf, get mixing, have low humidity, etc... and the cooling percentage grows quickly.
I managed a project to replace a failing UPS and it included UPS support for the air con.
There were two comms rooms each full of servers etc. having air con rated at 40kW max. Both rooms full to capacity. There was also a large office floor to support on UPS. Total load 160kW.
So 3 x 80kW UPS (giving N+1) were installed for the comms rooms and the office floor.
Then it was decided that the comms room air con must run when the servers, etc are on UPS, so a 4th 80kW UPS was installed. Thus the load of the 4th UPS was (1) 2 x 40kW air con and (2) one small air con unit in the UPS room (which housed 4 * 80kW UPS).
I do not recall how much overkill there was with the 4th 80kW UPS - it could be another 80kW UPS was purchased to match the first 3 and give future flexibility.
On the cooling side, and first of all, ten 20 ton CRACs is a bunch of cooling (703kW). Cooling is never 100% efficient, but way too much cooling for the heat load.
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A typical CRAC can have a coefficient of performance (COP) of 3. Thus it can deliver 3 watts of cooling for every watt of input power. To get 104 kilowatts of cooling, the CRAC system requires roughly 35kW of input power.
Thanks for the insight. When a manufacturer claims a CRAC is "20 ton", are they accounting for the heat generated by the CRAC itself (from that 35kw of input power)? If not, then the unit may provide 20 tons of cooling, but still put-back a third that in heat.
Good question and the following is only my understanding-- Liebert for example publishes a net capacity from the air. The gross load (going to a chiller for example) would be the net air plus the air handler input power. I bet the boundary conditions make it a loose number, but in general a 10 ton CRAC does not net 5 tons after it cools itself. It should be close to 10 depending upon room temps and humidity (and probably DX versus liquid rejection may change things slightly).
Thanks, DC. An answer I got from a Liebert VAR was that Liebert's published tonnage is "net" cooling, after subtracting heat generated by portions of the Liebert that are "within the airflow" (fan); but not devices outside the airflow (compressor). Their supposition was that the compressor cabinet can get nice and warm, but isn't vented, so doesn't exchange heat much with the room.
power usage for cooling versus UPSes
