# Air pressure drop off rate

I’ve been thinking about this for a while, and at first, I assumed that the inverse square law applied, but seeing as how air isn’t a form of radiation, I’m not sure that it still applies.

So if we have, let’s say, a fan, and the air density is $$\frac{1.225kg}{m^3}$$, then how fast would the fan’s air become stationary?

EDIT: Let’s assume it’s a 2 by 4 by 2 meter room, with smooth walls. There are no other air currents pressent, and the air temperature is normal room temperature. (22.5° C I believe). For the air’s velocity, I would prefer if there was a generalized formula for any speed, but if this is not possible, then assume it’s 10km/hr. The fan would just produce whatever wind currents a fan normally creates.

• How big is the room that the fan is in? What are the room's dimensions? What is the velocity of air exiting the fan? What is the air temperature? Are the walls smooth or rough? There are a LOT of details that would affect the answer. – David White Feb 28 '19 at 17:43
• @David White True, my mistake. Okay, let’s assume it’s a 20 by 4 room, with smooth walls. There are no other air currents pressent, and the air temperature is normal room temperature. (22.5° C I believe). For the air’s velocity, I would prefer if there was a generalized formula for any speed, but if this is not possible, then assume it’s 10km/hr. – Detmondyou Feb 28 '19 at 22:35
• Are your measurements in feet or meters? Is this the area of the floor, and if so, what is the height of the room? The devil is in the DETAILS. – David White Feb 28 '19 at 22:41
• The devilish details are not so much in the measurement scales, and only partially in the room shape (a tube-like room and a vast empty space of course change things relative to each other), but mainly in the actual flow field, turbulence and stuff like that. A fan that produced vortex rings would have a far bigger range than a fan that produced a lot of turbulence. – Anders Sandberg Feb 28 '19 at 23:17