We have a lot of the "silliest" answers here, from people who have no datacenter or AC experience. You put up a bounty, you get stupid spammers and beggars.
You basically have two choices:- Traditional refrigerant air conditioning with a condensing outdoor unit,
- Evaporative "Swamp coolers", where facilities allow large outside berth for building flow-through, and the local weather is favorable.
The amount of air conditioning required is calculable. You have two factors:
1. The amount of air conditioning required to keep an unloaded building at room temperature vs the hottest outside temperatures.
This is directly related to the building's insulation and R factor. If you have an uninsulated warehouse style steel building, you are going to be using much more AC to keep the building at room temperature than a highly insulated facility,
2. The amount of heat that needs to be removed from equipment heat generation.
I am not sure of the BTU rating, but I will need to dissipate upwards of 40,000 watts.
Unfortunately, #2 will be the major factor in designing an air conditioning system, your equipment-generated heat is much more than the amount needed for building cooling. The cons of air conditioning is that it is a closed system, so even on cool days, you'll be running AC equivalent to 40,000 watts of heat removal. This is one factor that has data centers looking for better ways of doing things.
Air conditioning has a lot of weird units of measurement, they can't seem to just use joules and watts like a normal physicist would. I will try to process some of these measurements like "tons" and "BTUs" to give you an idea about your AC power bills and required capacity.
ton = 12000 BTUs/hour, or 3517 watts. (based on how much ice would be used to provide the same refrigeration)
1 watt = 3600 joules per hour
1 btu = 1055.05585 joules
1 watt = 3.41214163 btu / hour
therm = 100,000 BTU
EER = Energy Efficiency Ratio = BTUs/watt-hour. BTU/hr vs watts of AC unit. A number 8-12 is typical
SEER = season-based voodoo. EER = -0.02 × SEER² + 1.12 × SEER
COP = Coefficient of perfomance. What we really want to know - i.e. how many watts will remove 1000 watt of heat. COP = EER / 3.412
The first thing to figure out is 40,000 watts equals how much in these AC terms, and how much electricity will it take. Lets remove everything except watts and the EER rating:
W
req = W
load / COP -> W
req = W
load * 3.412 / EER
So for 40,000 watts, and an example of 9 EER-rated air conditioning, we get
W
req = 40000W * 3.412 /
9 -> 15,164 watts
Next, how much AC capacity is required in those weird AC terms?
40000 watts = 11.4155251142 tons of air conditioning
So add that power use and capacity on top of what AC would normally be required for the space.
Evaporative cooling is measured a different way, in the temperature drop from intake air temp, with accompanying increased humidity. You can make 100F outside air into 75F inside air. However, you will need to look at the cubic feet per minute ratings of the systems to see what can keep up with your heat load. You may decide that 85F will be the maximum "output" temperature rise after air goes through your racks - for this much cooling, you will be looking at garage-door sized walls of fans from the outside and gallons of water per minute.
However, the evaporative cooling does have the advantage that you are putting in a massive outside air circulation system - the 75% of the day and year when outside air is below 75F, you will need nothing more than to run the fans.
Inside a closed air conditioned building, evaporative cooling may enhance efficiency a bit. AC removes humidity, to the point where the IDUs need to pump water out. You could add some humidity back to pre-cool the hot AC intake air (you can't humidify cold air AC output). The humidity would have to be strictly monitored to not go overboard or add more humidity than the AC can remove.
Whatever system is implemented, you need to direct airflow through your facility and systems, ideally in a typical contained hot/cool-aisle system:
