In short
- Heat load ≈ IT power + about 3 % for lighting, UPS and distribution losses in the room.
- Airflow: V = Q ÷ (ρ · cp · ΔT). At ΔT 12 K, 1 kW needs about 249 m³/h or 146 CFM.
- Water flow: ṁ = Q ÷ (cp · ΔT). At ΔT 10 K, 1 MW needs about 23.9 kg/s, roughly 1,435 l/min.
- 1 refrigeration ton = 3.517 kW; 1 MW is about 284 tons or 3.41 million BTU/h.
- Doubling ΔT halves the flow, on the air side and on the water side.
Heat load: start from IT power
Servers convert practically all of their electrical input into heat, so the IT load in kW is the heat load in kW. Add the other heat in the room:
- UPS and PDU losses if they sit in the white space (often 2–5 % of the load they carry).
- Lighting, people and the building envelope: small in a large hall, often 1–2 % together.
For screening, IT plus 3 % is a reasonable figure. Use the design IT load, not today's, and decide whether you size for nameplate or for measured peak; most designers use the contracted rack power.
Airflow for the heat that goes to air
| ΔT across IT | m³/h per kW | CFM per kW |
|---|---|---|
| 10 K (18 °F) | 298 | 176 |
| 12 K (21.6 °F) | 249 | 146 |
| 15 K (27 °F) | 199 | 117 |
| 20 K (36 °F) | 149 | 88 |
In imperial units the same rule is CFM = 3.16 × W ÷ ΔT(°F). Servers set their own ΔT through their fans, typically 10–20 K; the cooling units must supply at least the air the IT pulls, plus a margin for leakage.
Water flow for the heat that goes to liquid
With PG25 (25 % propylene glycol, cp about 3.85) the same duty needs about 1,510 l/min, around 5 % more, and the pump works harder because the fluid is more viscous. Glycol is used where the loop can freeze, typically on the dry-cooler side.
Converting to tons and BTU/h
| From | To |
|---|---|
| 1 kW | 3,412 BTU/h |
| 1 refrigeration ton (RT) | 3.517 kW = 12,000 BTU/h |
| 1 MW | ≈ 284 RT |
| 40 kW rack | ≈ 11.4 RT |
A hall of 40 GB200-class racks
40 racks at 132 kW, 85 % of the heat captured by direct-to-chip liquid, 3 % other room heat, air ΔT 12 K, water ΔT 10 K:
| Quantity | Value |
|---|---|
| IT load = 40 × 132 kW | 5,280 kW |
| Total heat = 5,280 × 1.03 | 5,438 kW ≈ 1,547 RT |
| To liquid = 85 % of IT | 4,488 kW |
| Water flow = 4,488 ÷ (4.18 × 10) | 107 kg/s ≈ 6,440 l/min |
| To air = 15 % of IT + room heat | 950 kW |
| Airflow = 950 × 249 m³/h | ≈ 236,000 m³/h · 139,000 CFM |
Even at 85 % liquid capture the hall still needs almost a megawatt of air cooling. That residual is why liquid-cooled halls keep fan walls or CRAHs.
Sizing the plant
- Design outdoor condition: size heat rejection for the site's extreme dry-bulb (or wet-bulb for towers) from ASHRAE climatic data, not the average day.
- Redundancy: add N+1 or 2N on chillers, dry coolers, pumps and CDUs to match the tier you are building.
- Free-cooling limit: facility water supply minus the dry-cooler approach. W32 with a 6 K approach gives full free cooling up to about 26 °C outdoor; see the free-cooling chain.
- Water temperature: pick the warmest class the IT allows; the ASHRAE limits list them.
Questions about cooling calculations
How many CFM per kW does a server rack need?
About 146 CFM per kW at a 12 K (21.6 °F) rise, or 158 CFM per kW at 20 °F. A 20 kW rack therefore needs roughly 2,900–3,200 CFM.
How many tons of cooling for a 1 MW data center?
About 284 refrigeration tons for the IT alone, around 293 tons with 3 % other room heat, before redundancy.
Should I include latent heat?
Data halls have almost no moisture sources, so the load is close to 100 % sensible. Latent load matters mainly for fresh-air treatment.
Why does ΔT matter so much?
Flow is inversely proportional to ΔT. Running 15 K instead of 10 K on the water side cuts the flow by a third, allows smaller pipes and lowers pump energy by even more.
Sources
- Air density 1.20 kg/m³, cp 1.006 kJ/kg·K; water cp 4.18 kJ/kg·K; PG25 cp about 3.85 kJ/kg·K.
- ASHRAE TC 9.9 thermal guidelines; ASHRAE Handbook, Fundamentals (climatic design data).
- Worked figures are screening calculations, not a design.