# How can I calculate the airflow needed for a fan to cool a system?

The airflow needed for a system to sustain itself at 80°C with 25°C ambient temperature can be calculated with this formula:

Airflow = (P * t) / (ΔT * D * SHC) [m3/s]


Where

P = Power [watts]

t = Time [seconds]

ΔT = Difference in Temperature [°C or K]

D = Density of Air [kg/m3]

SHC = Specific Heat of Air [J/(kg*K)]


For a 105W CPU to remain at 80°C the airflow needs to be:

Airflow = (105 * 1) / (55 * 1.2 * 1000)

Airflow = 0.00159m3/s or 5.27m3/h


This means any small 80mm fan typically used for case mods to improve airflow by a little:

Can cool a 3900X beast CPU that requires an air cooler with a beefy heatsink and a large fan:

How can I further improve this formula to take into account the thermal resistance of the heat source and some fluid dynamics inside the case when it's simplified to just a box?

I know this is more of a thesis amount of work that's needed to be done but I'm only looking at a very rough estimate.

• °C or °F? Hit the edit link. May 2 '20 at 13:19
• Updated the post with calculations. May 2 '20 at 13:20
• Watch the other units too. The kilogram is 'kg'. A 'Kg' is a Kelvin-gram which is nonsense. You can use <sup>...</sup> and <sub>...</sub> for superscript and subscript for your 'cubed'. May 2 '20 at 13:23
• It's J/kg*K, here: socratic.org/questions/what-is-the-specific-heat-of-air May 2 '20 at 13:29
• Also, the temperature difference should have no unit attached to it. May 2 '20 at 13:35

I usually work out the equation from first principles to solve for t.

$$t = \frac {\Delta T \times m \times SHC} P$$

where t is in seconds, ΔT is in K or °C, m is in kg and SHC is J/kg·K.

= SHC of Air ~= 1000 J/kg·K. - Density of Air ~= 1.2 kg/m3. - Airflow ~= 0.035 m3/s. - Mass flow = airflow × density = 0.035 × 1.2 = 0.042 kg/s - Power ~= 265 watts

Temperature difference = Power / (Specific Heat of Air * Density of Air * Airflow)

Rearranging we get

$$\Delta T = \frac {P \times t} { m \times SHC} = \frac {265 \times 1} { 0.042 \times 1000} = 6.3° \text C$$

This tells us that if the temperature of the PC is stable then the air will exit the box 6.3°C warmer than it goes in. Does this match your experimental readings?

You can rearrange the equation to calculate mass flow for a particular ΔT.