Acceleration of Airparticles in Cyclone?

Question

What would the equation be to determine the acceleration of air entering a cyclone? If possible i would like to also know how different air pressures relates to that acceleration. So, say the air pressure entering is 1 atm, and the cyclone is a vaccume. How would the acceleration and pressure of the air entering change?

Answer 1

The force in a cyclone is similar to a buoyant force. Although a buoyant force always acts upwards, it is caused by a pressure differential. Forces due to pressure gradients depend only on the volume being acted upon, apart from whatever relation governs the differential itself.

Standard atmospheric pressure is defined as 101,325 Pa (1013 mbar). In a very powerful hurricane or typhoon, the minimum central pressure may be 90,000 Pa (900 mbar), and such a powerful storm could easily have a radius of 200 km. The pressure gradient would therefore be about 10,000 Pa / 200 km = (10000 N/m^2) / (200000 m) = 0.05 N/m^3. (Saying that the cyclone is a vaccum is not a reasonable assumption.)

This means that a cubic meter of air in the storm would have 0.05 N acting on it, towards the center of circulation. The average density of air at sea level is 1.2 kg/m^3, so that same cubic meter of air has a mass of 1.2 kg. As a result, the acceleration of the air should be 0.05 N / 1.2 kg = 0.00417 m/s^2. Not a large number at all. But let's see what happens when we apply it across the entire 200 km travel path. I'll use the formula v = sqrt(2ax) = sqrt(2*0.00417*200000) = 40.84 m/s. Converting that to a more typical measure of wind speed gives us (40.84 m/s)(1 km / 1000 m)(3600 s / 1 hr) = 147 km/hr (91 mi/hr). That's a bit lower than the wind speed would normally be for a storm as powerful as I described, but it's in the right ballpark at least. I would guess that the extra wind speed comes from phase changes in water vapor and from the movement of the storm center itself, among other meteorological processes.