Network & Power
Rack heat load
Equipment watts → BTU/hr and the cooling airflow the rack room needs.
Every watt an amplifier, switch, or processor draws ends up as heat in the closet. Add the gear (or a total wattage), get BTU/hr, tons of cooling, and the CFM of airflow required for a target temperature rise. Amplifiers get a duty-cycle factor because they rarely run at rated output.
gear in the rack
Enter load as
heat 60 W
heat 44 W · 125 W out × (1/0.90 − 1) + 30 W idle
heat 157 W · 120 W + 10% of 370 W PoE; the rest leaves on the cables
Heat load
890BTU/hr
261 W of heat · 0.07 tons of cooling
W × 3.412 = BTU/hr · ÷ 12,000 = tons · amps at 1/8 duty, 90% efficient
Airflow for 15 °F rise
55CFM
93 m³/h
BTU/hr ÷ (1.08 × 15) — free-air rating; fan trays deliver 40–60% of nameplate once filters and rack impedance are counted
- ▸ Size rack fans at roughly 2× the CFM above to cover filter loading and back-pressure. Pull hot air out at the top; let cool air in low at the front. Blank unused rack units so air goes through gear, not around it.
- ▸ PoE power leaves the rack with the cable and becomes heat at the camera or AP. Only the switch's own draw plus ~10% PSE conversion loss stays in the closet.
- ▸ Amplifier heat is input minus acoustic output. A 4 × 250 W class-D amp at music duty is closer to 45 W of heat than 1000 W — size the room, not the nameplate.
- ▸ If the closet has no mechanical cooling, this heat raises the room temperature until the walls and door shed it. Above ~1000 BTU/hr in a small closet you need a duct or a split unit, not a fan.
heat by device
Each slice is one row's contribution to the total heat staying in the rack. The arrow scales with the airflow needed to carry it away.
Need this math wired into a real system? We do that for a living.
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