Data center cooling calculation: kW per rack to CFM and tons

What this covers

• Where the 0.316 and 3.16 factors come from and when they stop being accurate.
• How the server temperature rise changes CFM per kW, with a curve from 10 to 35°F.
• A worked room with 8, 20, and 40 kW racks carried from IT load to tons.
• How bypass air, UPS losses, and cooling unit fans change the total, and what N+1 and 2N add.

What changes the result

• The temperature rise across each group of servers, set by the servers' own fans and firmware.
• Bypass air that leaves cooling units and returns without passing through any IT equipment.
• Non-IT heat in the room, including UPS losses, lighting, and the cooling units' own fans.
• The redundancy level, which changes the installed unit count without changing the load.

Every watt of IT load becomes a watt of heat

Servers, storage, and network equipment do no external work, so the electrical power they draw leaves them as heat in the air passing through them. The cooling load of an air-cooled IT room is therefore its measured or planned IT power, plus every other source of heat in the room, and none of it is latent except what people and outdoor air bring.

Use the power the equipment will draw at full build, from metering or from the maker's configuration power figures, not the nameplate rating of each power supply. Power supplies are rated for the largest draw they can deliver, and summing nameplates inflates the load, the airflow, and the plant.

Where 0.316 and 3.16 come from

For standard air, sensible heat in BTU per hour equals 1.08 × cfm × ΔT, where 1.08 is 60 minutes times an air density of 0.075 pounds per cubic foot times a specific heat of 0.24 BTU per pound per °F. One watt is 3.412 BTU per hour, so dividing 1.08 by 3.412 gives watts = 0.316 × cfm × ΔT.

Rearranged, cfm = watts ÷ (0.316 × ΔT), which is 3.16 × watts ÷ ΔT. The same relationship in SI units is watts = 1.2 × 1,006 × airflow in cubic meters per second × ΔT in kelvin, about 1,210 × m³/s × ΔT.

The factor assumes air near sea level and room temperature. At high altitude the air is thinner, each cubic foot carries less heat, and the airflow for the same load rises. At 5,000 feet air density is about 17 percent below sea level, so the 3.16 factor becomes about 3.8. Correct the density for sites well above sea level.

CFM per kW depends on the server temperature rise

Per kilowatt, the formula gives 3,160 ÷ ΔT cfm. The temperature rise is what varies: a server that heats its air by 20°F needs 158 cfm per kW, and one that heats its air by 30°F needs 105. The rise belongs to the servers, whose fans respond to their own inlet temperature and processor load, not to the cooling units.

That has a design consequence. A cooling unit selected for a large return-to-supply temperature difference cannot force servers to run a large rise; if the servers pull more air than the units supply, the shortfall is made up by hot exhaust air recirculating to server inlets. Size airflow from the servers' own rise at the design inlet temperature.

Worked example: a room with three rack densities

The room holds 40 racks at 8 kW, 10 racks at 20 kW, and 4 racks at 40 kW, 680 kW of IT load. Assume the 8 kW racks run a 20°F rise, the 20 kW racks 25°F, and the 40 kW racks 30°F. An 8 kW rack needs 3.16 × 8,000 ÷ 20 = 1,264 cfm, a 20 kW rack 3.16 × 20,000 ÷ 25 = 2,528 cfm, and a 40 kW rack 3.16 × 40,000 ÷ 30 = 4,213 cfm.

By group, that is 50,560, 25,280, and 16,850 cfm, 92,690 cfm in total. Averaging the load first and applying one rise would give a different and wrong number, because airflow scales with each group's own rise. Averaging all 680 kW at 20°F, for example, gives 107,440 cfm of IT airflow, 16 percent more than the groups actually need.

The four 40 kW racks may not be air-cooled at all in a real room. Where they use rear door heat exchangers or direct liquid cooling, subtract the share of heat the maker states goes to water, and carry only the remainder through this calculation.

Bypass air, recirculation, and containment

Cooling units have to supply more air than the servers draw. Some supply air leaks through cable openings, around racks, and into empty rack positions and returns to the units without passing through any equipment. The example allows 15 percent of IT airflow for bypass, 13,900 cfm, bringing supply to 106,600 cfm.

Bypass is a room condition, not a constant. Hot aisle or cold aisle containment, blanking panels, and sealed floor openings cut it, and each percentage point removed is fan energy no longer spent. The opposite failure, supplying less air than the servers draw, sends hot exhaust back to server inlets and raises inlet temperatures even though the cooling units report spare capacity.

Non-IT heat and the cooling units' own fans

Heat from equipment that is not IT load still has to be removed if it sits in the cooled space. The example assumes a UPS inside the room losing 4 percent of 680 kW, 27.2 kW, plus 3.0 kW of lighting and 2.0 kW of envelope gain, 32.2 kW together.

Cooling unit fans add heat to the air they move. At an assumed 0.3 watts per cfm on 106,600 cfm, the fans add 32.0 kW, which the units must remove on top of everything else. Use the unit maker's fan input at the operating airflow and pressure for a real selection, since fan heat is the one load that grows with the airflow the calculation just produced.

The room total is 680 + 32.2 + 32.0 = 744.2 kW. That fan heat is the same effect the cleanroom heat load calculation counts for fan filter units, and for the same reason: fan power becomes heat inside the conditioned space.

Converting to tons and counting units

One ton of refrigeration is 12,000 BTU per hour, and one kilowatt is 3,412 BTU per hour, so tons = kW × 0.2843. The room needs 744.2 × 0.2843 = 211.6 tons of sensible cooling, or 2,539,000 BTU per hour.

With cooling units rated at 100 kW of net sensible capacity at the design return condition, N is 744.2 ÷ 100 rounded up, 8 units. N+1 installs 9, so any one unit can fail or be serviced with the full load still covered. 2N installs 16 on two independent systems, so a whole system can be lost. Each running unit in the N case moves 13,325 cfm and removes 93 kW, a 22°F drop across its coil, and its expanded performance data has to confirm that capacity at the actual return air temperature. The redundancy design page compares these options, including running all 9 units at part speed.

Checking the result against the thermal envelope

The calculation holds only if server inlet air stays inside its allowed range. ASHRAE's thermal guidelines recommend inlet air of 18 to 27°C (64.4 to 80.6°F) with a dew point between minus 9 and 15°C, and wider allowable ranges by equipment class. The ASHRAE contamination guideline adds that relative humidity should stay below 60 percent so settled dust does not absorb enough moisture to become conductive, and recommends the room be kept clean to ISO 14644-1 Class 8.

That cleanliness recommendation is where data center and cleanroom design meet, and the cleanroom inside a data center case study follows a room that has to satisfy both at once.

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