-1
$\begingroup$

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:

enter image description here

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

enter image description here

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.

$\endgroup$
7
  • $\begingroup$ °C or °F? Hit the edit link. $\endgroup$
    – Transistor
    May 2 '20 at 13:19
  • $\begingroup$ Updated the post with calculations. $\endgroup$
    – Vulkan
    May 2 '20 at 13:20
  • $\begingroup$ 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'. $\endgroup$
    – Transistor
    May 2 '20 at 13:23
  • $\begingroup$ It's J/kg*K, here: socratic.org/questions/what-is-the-specific-heat-of-air $\endgroup$
    – Vulkan
    May 2 '20 at 13:29
  • $\begingroup$ Also, the temperature difference should have no unit attached to it. $\endgroup$
    – Vulkan
    May 2 '20 at 13:35
1
$\begingroup$

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.

In your case you've got:

= 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.


From the comments:

Also, the temperature difference should have no unit attached to it.

That's not true. The difference between any two measurements will have the units of the original measurements. The difference between two lengths will be in metres, the difference between two times will be in seconds and the difference between two temperatures will be in °C or K (or °F if you're from some of the British colonies).

$\endgroup$
3
  • $\begingroup$ So, in order to keep it under 90°C I will need to calculate for a ΔT = 65°C. The answer doesn't take into consideration the volume of the case. $\endgroup$
    – Vulkan
    May 2 '20 at 15:34
  • $\begingroup$ I think considering the case volume makes things a lot more complex. $\endgroup$
    – Vulkan
    May 2 '20 at 15:45
  • $\begingroup$ Where did you get those numbers? I sounds as though you think that it will run at 155°C with no fan. To model this properly you'd need to know the thermal resistance from the heat source through the heatsink, etc. I don't think I can help further. $\endgroup$
    – Transistor
    May 2 '20 at 15:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.