Assuming constant fan power, how would atmospheric pressure/density affect the cooling rate of objects being cooled by a fan? Like in a lower pressure atmosphere, would a fan be able to cool objects more quickly because it could blow less dense air at a faster speed, would it cancel out and be the same or would the cooling rate be lower under less atmospheric pressure/density?

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    $\begingroup$ Ask yourself this question: What would happen if the air density were zero? $\endgroup$ Commented May 30, 2017 at 0:40
  • $\begingroup$ I'll add to Chester Miller's comment. Please provide more information. Is the object outdoors or indoors? Regarding heat transfer, you will normally find that convection, conduction, and radiation, are ALL happening at the same time. If you don't specify more constraints on the problem, you will not get a reasonable answer. $\endgroup$ Commented May 30, 2017 at 1:31
  • $\begingroup$ Well I'm thinking of an object that's inside an air tight container. $\endgroup$
    – Tom
    Commented May 30, 2017 at 1:44
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    $\begingroup$ If the air tight container has vacuum in it (zero density), the fan blades will just spin, but will not push any air around, so the fan won't contribute to any cooling whatsoever. $\endgroup$ Commented May 30, 2017 at 12:36
  • $\begingroup$ Right, and I can imagine that the cooling ability of a fan approaches zero as the density of air does, but what I was wondering was as the density of air approaches zero, at what rate does the cooling decrease? Would it just be a linear decrease, or is it otherwise? Like if the mass of air being blown is lower by a certain proportion, but the velocity is higher by the square root (because power of the fan is constant), and the cooling rate would decrease by the square root of the air density decrease. $\endgroup$
    – Tom
    Commented May 30, 2017 at 13:26

2 Answers 2


This can be understood using heat transfer formulae $$dQ=h \cdot A \cdot (T_{air}-T_{solid})$$ where heat transfer coefficient is an empirical function of velocity and density. So to remove heat, one need to either increase velocity or increase density. In your question, you increase velocity and reduce density, so the trend is uncertain.

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    $\begingroup$ Increase velocity and reduce "density"... not speed. $\endgroup$
    – Rob S.
    Commented Jun 5, 2017 at 21:19

No, it can't.

Just consider this: Some unmanned aircraft have pressurized compartments in their fuselage. Sounds strange? If no pilot is on board, why bother with pressurization?

The low air density at altitude would not be sufficient to cool the electronics, so a pressurized electronics compartment had to be added. If you keep the fan power constant, it would only increase flow speed by the square root of the density difference, leaving you with a reduced cooling performance.

  • $\begingroup$ That does answer my question. I have to ask, does this mean that with the mass of air being blown proportionally decreasing with pressure and velocity increasing proportional to the square root, does that mean the effectiveness of the fan approaches zero but does so by the square root of the air pressure? $\endgroup$
    – Tom
    Commented May 31, 2017 at 16:32
  • $\begingroup$ Define effectiveness of the fan first! If you adjust the fan speed to the flow speed, its efficiency will remain unchanged. $\endgroup$ Commented May 31, 2017 at 18:29

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