Rack power density calculator
A room has a load. A rack has a limit. The gap between them is the floor plan.
Enter a total IT load and how many cabinets it is spread across. The calculator returns density per rack and per rack unit, the current each feed carries, the breaker that current implies once the continuous-load rule is applied, and the airflow the rack needs at a typical temperature rise.
Density is where capacity planning usually goes wrong, because a room is rarely limited by its total kilowatts. It is limited by what a single cabinet can be fed, cooled and physically supported — and those three limits are almost never the same number.
Worked example — 250 kW across twenty cabinets
A mid-sized hall at a density that still works on contained room air.
- Total IT load — 250 kW. Measured or forecast IT power for the whole room, at the rack boundary.
- Populated cabinets — 20 racks. Cabinets actually carrying IT load. Empty cabinets do not dilute density.
- Usable height per cabinet — 42 U. Rack units available for equipment after PDUs, patching and blanking.
- Rack supply — 230 V single phase. Voltage and phase configuration delivered to the rack PDU.
- Air temperature rise — 11 °C. Difference between cabinet inlet and exhaust, used only for the airflow estimate.
Power per cabinet: 12.50 kW — High density — needs containment, managed inlet temperature and per-rack monitoring
| Watts per rack unit 42 usable U per cabinet | 298 W/U |
|---|---|
| Current if one feed carries the rack 230 V single phase | 54.3 A |
| Current per feed at a balanced A/B draw What you see day to day — not what the feed must be sized for | 27.2 A |
| Breaker implied by the continuous-load rule Load is 67.9 A at 125% | 80 A |
| Airflow required per cabinet 3,392 m³/h at ΔT 11.0 °C | 1,997 CFM |
| Cabinets needed at 5 kW each | 50 racks |
| Cabinets needed at 15 kW each | 17 racks |
Above roughly 12 kW a cabinet stops being an airflow problem the room can absorb and becomes one the row has to be designed around. Check the aisle before ordering the PDU.
Formula
kW per rack = total IT load ÷ populated cabinets; amps = kW × 1000 ÷ V (÷ V × √3 for three phase)
The breaker figure applies the continuous-load rule: a circuit feeding a rack is sized at 125% of the load, which is the same as loading it to no more than 80%. Airflow uses the same air-property constants as the cooling load calculator.
Density bands and what they demand
Rules of thumb for orientation. The binding constraint at any density is whichever of feed, cooling and floor loading runs out first.
| Per cabinet | Commonly described as | What it usually requires |
|---|---|---|
| 1 – 3 kW | Legacy room | Open floor, minimal containment, generous aisle width |
| 3 – 6 kW | Standard enterprise | Blanking panels, brush grommets, disciplined cable management |
| 6 – 12 kW | Modern colocation | Hot or cold aisle containment, per-rack metering |
| 12 – 25 kW | High density | Full containment, managed inlet temperature, in-row or close-coupled cooling |
| 25 – 50 kW | Very high density | Rear-door heat exchangers or direct-to-chip capture for part of the load |
| Above 50 kW | AI and HPC | Liquid cooling, reinforced floor loading, dedicated distribution |
Average density is a planning fiction
Dividing a room load by a cabinet count produces an average, and almost no real hall is uniform. A few cabinets full of GPUs sit next to rows of network gear drawing a tenth as much, and the cooling system has to survive the peak cabinet, not the average one. Use this number to size the room and then check the worst cabinet separately.
The same applies to the feed. A rack PDU rated at 32 A does not mean 32 A is usable: the continuous-load rule caps it at about 80%, and a dual-fed cabinet has to survive losing one side, which means each feed carries the whole cabinet rather than half of it. Planning against the balanced day-to-day figure is how a maintenance window becomes an outage.
Three limits, rarely the same number
- Electrical. What one circuit can deliver at the installed voltage, after the continuous-load derate and with one feed out of service.
- Thermal. What the row can actually take away, which depends on containment, inlet temperature and the airflow the cabinet can pull through itself.
- Physical. Cabinet depth, weight on the floor or raised-floor tile, cable bend radius and the space in front for a person to work safely.
Why watts per U matters more than it used to
A 42U cabinet at 5 kW spends about 120 W per rack unit, which almost any air-cooled design absorbs. The same cabinet at 40 kW spends nearly a kilowatt per rack unit, and that heat has to cross a few centimetres of chassis before it reaches any air at all.
This is the mechanism behind the shift to liquid: not that air stops working in the room, but that it stops working inside the box. Watts per U is the earliest warning that a design is heading toward rear-door exchangers or direct-to-chip capture.
Frequently asked questions
How many kW per rack is normal?
Most enterprise rooms run 3 to 6 kW per cabinet, modern colocation halls run 6 to 12 kW, and high-density or AI deployments run well above 20 kW. The average across a whole facility is usually much lower than its busiest cabinets, which is why cooling is designed around the peak rack rather than the mean.
How many amps does a 5 kW rack draw?
At 230 V single phase, 5 kW draws about 21.7 A. At 208 V single phase it is about 24 A, and at 400 V three phase about 7.2 A per phase. Apply the continuous-load rule before choosing a breaker: a 21.7 A continuous load wants a 32 A circuit, not a 25 A one.
Why does each feed have to carry the whole rack?
Because the point of a second feed is surviving the loss of the first. If A and B each carry half the cabinet and A is switched off for maintenance, B is suddenly at 100% — and if it was sized for 50%, it trips. Dual-corded equipment is normally planned so either feed alone can carry the full load.
What is watts per U and why does it matter?
It is the power a cabinet dissipates divided by its usable rack units, and it describes how concentrated the heat is rather than how much of it there is. Around 900 W per U and above, the constraint moves from the room to the chassis itself, which is where rear-door exchangers and direct-to-chip cooling start to appear.
How do I convert kW to amps for a three-phase rack?
Divide the load in watts by the line voltage multiplied by the square root of three. For 10 kW at 400 V three phase: 10,000 divided by (400 × 1.732) is about 14.4 A per phase, assuming a balanced load and a power factor near one.
Does an empty cabinet lower my density?
It lowers the facility average and changes nothing about the cabinets that are loaded. Density planning should count populated cabinets; counting empty ones produces a comfortable number that the cooling system will not recognise.
Where this calculation stops
This is arithmetic at the rack boundary. It does not size a distribution board, check selectivity or protection coordination, account for power factor and harmonics, or verify that the floor can carry the cabinet.
The airflow figure assumes standard air near sea level and a uniform temperature rise. Real cabinets recirculate, leak and stratify, and a CFD model exists precisely because that arithmetic is not the whole story.