Enclosure Fan Airflow Calculator (CFM)
Enclosure Fan Airflow Calculator (CFM)
How much air a sealed-but-vented enclosure needs to carry its own heat away: Q = P ÷ (ρ·c_p·ΔT), derived rather than quoted, with the altitude and inlet-temperature corrections that the familiar 3.16·W/ΔT°F shortcut leaves out — and the fan-curve derating that is the reason a fan never delivers its rated CFM.
Enclosure airflow
300 W of losses inside a panel, a 10 °C rise allowed, 35 °C inlet air at sea level, a fan expected to reach 60% of its free-air rating with another 15% lost to the filter, and a 150 CFM fan in mind
Where 3.16 and 1.76 come from
ρ·cp = 1.2 × 1005 = 1,206 J/(m³·K) at 21 °C and sea level
Q[CFM] = 2,118.88 ÷ 1,206 × P/ΔT°C = 1.757·P/ΔT°C
= 1.757 × 1.8 × P/ΔT°F = 3.162·P/ΔT°F
Q[m³/h] = 3,600 ÷ 1,206 × P/ΔT°C = 2.985·P/ΔT°C
ρ = p ÷ (R·T), p = 101,325·(1 − 2.25577×10⁻⁵·h)5.25588 (ISO 2533)
- rho c p
- the heat one cubic metre of air carries per degree. Everything about enclosure cooling is this number
- delta T
- the rise of the AIR between inlet and outlet, not of any component in it
- h
- altitude. Pressure falls, density falls with it, and the airflow needed rises in exactly the same proportion
- Q free
- a fan’s catalogue CFM, measured against no back pressure. The enclosure never lets it deliver that
Worked example
300 W of losses inside a panel, a 10 °C rise allowed, 35 °C inlet air at sea level, a fan expected to reach 60% of its free-air rating with another 15% lost to the filter, and a 150 CFM fan in mind
At 35 °C and sea level the air is 1.1455 kg/m³, so ρ·cp = 1,151.2 J/(m³·K) — 4.5% less than the 1,206 the shortcut assumes, because the air is hot
Q = P ÷ (ρ·cp·ΔT) = 300 ÷ (1,151.2 × 10) = 0.02606 m³/s = 55.2 CFM = 93.8 m³/h
The 1.757·W/ΔT°C shortcut would have said 52.7 CFM — 4.8% low, and that is only the inlet temperature; add altitude and the gap widens
The operating point and the filter together deliver 51.0% of a fan's rating, so the fan has to be rated 55.2 ÷ 0.5100 = 108.3 CFM free air
The 150 CFM fan therefore delivers about 76.5 CFM here, which carries 416 W at a 10 °C rise — enough, and the rise it actually gives at 300 W is 7.2 °C
So the air inside settles at about 42.2 °C. Every component inside sits its own junction-to-air rise above that
The altitude correction, checked against a published fan-engineering table
| Altitude | Density ratio (this page) | Airflow multiplier | Aerovent FE-1600 factor |
|---|---|---|---|
| Sea level | 1.0000 | 1.000 × | 1.00 |
| 457 m | 0.9470 | 1.056 × | 1.06 |
| 914 m (about Bengaluru) | 0.8962 | 1.116 × | 1.12 |
| 1,524 m | 0.8320 | 1.202 × | 1.20 |
| 2,134 m (about Shimla or Abha) | 0.7716 | 1.296 × | 1.30 |
| 3,048 m | 0.6877 | 1.454 × | 1.45 |
The shortcut forms, and exactly what they assume
| Form | Units | Assumes | Equivalent to |
|---|---|---|---|
| Q = 3.16 × W ÷ ΔT°F | CFM | 21 °C air at sea level | 2,118.88 ÷ 1,206 × 1.8 |
| Q = 1.76 × W ÷ ΔT°C | CFM | the same | 2,118.88 ÷ 1,206 |
| Q = 3.0 × W ÷ ΔT°C | m³/h | the same | 3,600 ÷ 1,206 |
| Q = 0.05 × W ÷ ΔT°C | m³/s | the same | 1 ÷ 1,206 × 60 |
One stream, one energy balance, and the two things that spoil it
A vented enclosure in steady state is the simplest heat exchanger there is. Every watt dissipated inside has to leave in the air, so the heat carried, ṁ·cp·ΔT, equals the heat put in. Volumetric flow is mass flow over density, and that is the whole derivation: Q = P/(ρ·cp·ΔT). Fill in 1.2 kg/m³ and 1,005 J/(kg·K) and the familiar shortcuts appear — 1.76·W/ΔT°C and 3.16·W/ΔT°F in CFM, about 3.0·W/ΔT°C in m³/h. They are not separate rules of thumb; they are that one equation with ρ·cp = 1,206 J/(m³·K) folded in, and they carry that assumption whether you know it or not.
Thin air carries less heat. ρ·cp is joules per cubic metre per degree, so if the air is less dense — because it is hot, or because you are up a hill — each cubic metre carries less and you need more of them. The two effects multiply. A panel at 45 °C in Bengaluru at 920 m is working with air about 15% thinner than the shortcut assumes, and needs about 15% more flow than it says. This page takes the pressure from the ISO 2533 standard atmosphere and the density from the ideal gas law at your actual inlet temperature, and the altitude half of that model agrees with the published fan-engineering correction tables to better than half a per cent.
A fan’s rated CFM is a figure it will never give you. Catalogue airflow is measured in a test chamber against no back pressure at all — free air, to ANSI/AMCA 210. A real enclosure resists the flow, and that resistance rises roughly as the square of the flow: it is a parabola through the origin on the same pressure-versus-flow axes the fan’s curve is drawn on. The fan delivers whatever flow puts its curve and that parabola at the same point, and that operating point is always down and to the left of the free-air figure. Half to seventy per cent is a fair expectation for a reasonably open box; a crowded enclosure with fine louvres and a filter can be much worse, and a filter that has been in service for a year is worse again. That single fact explains most of the fans that were correctly sized on paper and are not coping in the field. If the enclosure is restrictive, two fans in series raise the available pressure and move the operating point back to the right; two in parallel raise the free-air flow and barely help.
What this page does not tell you. It gives the temperature of the air, not of anything in it. A MOSFET on a heatsink in that enclosure is still its own junction-to-air thermal resistance above the internal air temperature, which is what the heatsink thermal resistance calculator is for — and the ambient it needs is the internal air temperature this page gives, not the room’s. It also takes no credit for heat conducted or radiated through the enclosure walls, which for a metal cabinet in still air is real but modest, so the answer errs on the generous side. And it assumes the air is well mixed: a fan that short-circuits from inlet to outlet across one corner can leave a dead pocket 20 °C hotter than this calculation says, which no amount of extra CFM will fix. Put the inlet low and the outlet high, and make the air go past the hot parts.
The losses that make up the watts come from the MOSFET loss, diode loss and LDO loss pages; the BTU to watts converter turns an air-conditioning figure into the same units if the cabinet ends up needing one.
Frequently asked questions
How do you calculate the CFM needed to cool an enclosure?
Q = P ÷ (ρ·cp·ΔT). With standard air at 21 °C and sea level that is 1.76 × watts ÷ ΔT°C, or 3.16 × watts ÷ ΔT°F, in CFM. Both constants are 2,118.88 CFM per m³/s divided by ρ·cp = 1,206 J/(m³·K), so they are only right for that air.
Where does the 3.16 in the fan airflow formula come from?
From ρ·cp. One cubic metre of standard air carries 1,206 joules per degree; there are 2,118.88 CFM in a cubic metre per second; and a Fahrenheit degree is 1/1.8 of a Celsius one. 2,118.88 ÷ 1,206 × 1.8 = 3.162. It is not a measured constant, it is that arithmetic.
Do I need more airflow at altitude?
Yes, in inverse proportion to the air density. At 920 m the air is about 10% thinner than at sea level at the same temperature, so you need about 11% more volumetric flow for the same heat and the same rise. At 2,200 m it is nearer 30%. The fan’s pressure capability falls in the same proportion, so the operating point moves too.
Why does my fan not deliver its rated CFM?
Because the rating is the free-air figure, measured with no back pressure, to ANSI/AMCA 210. In an enclosure the fan settles where its pressure-flow curve crosses the enclosure’s system impedance — a parabola rising as the square of the flow — and that is always at a lower flow. Expect 50–70% of the rating in an open box and less behind a filter.
Should the fan blow in or suck out?
Blowing in pressurises the enclosure, so every cubic metre entering goes through the filter and dust cannot come in through the seams; the fan also runs in cooler air, which is better for its bearings. Sucking out gives a slightly lower internal temperature at the fan itself but draws unfiltered air in through every gap. For dusty environments, blow in.
Is this the temperature my components will reach?
No — this is the temperature of the air inside. Each component then sits its own junction-to-air thermal resistance above that. Use the internal air temperature this page gives as the ambient for a heatsink calculation, not the room temperature.
Related calculators
References
- Aerovent (Twin City Fan), Temperature and Altitude Effects on Fans, Fan Engineering bulletin FE-1600 (metric edition). States the fan industry’s standard air as 1.2 kg/m³ at 21 °C, sea level, 101.32 kPa; gives the constant-speed fan laws (volume unchanged, pressure and power varying with density ratio); and publishes the altitude correction table this page’s model is checked against, eight rows from sea level to 3,048 m.
- ISO 2533:1975, Standard Atmosphere. The troposphere pressure relation p = p₀(1 − 2.25577×10⁻⁵·h)^5.25588 used here for the altitude correction, valid to 11,000 m.
- ANSI/AMCA 210-16 and ANSI/ASHRAE 51-16, Laboratory Methods of Testing Fans for Certified Aerodynamic Performance Rating. The standard behind the numbers on a fan’s data sheet, and the reason a catalogue CFM is a free-air figure measured in a chamber rather than anything you will see in an enclosure.
- ASHRAE Handbook — Fundamentals, chapter on the thermodynamic properties of moist air, for cp ≈ 1,005 J/(kg·K) for dry air near room temperature and the ideal-gas density relation ρ = p/(R·T) with R = 287.05 J/(kg·K). Used to derive ρ·cp rather than to look up an airflow constant.