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December 24

How does funnels affect wind speed?

I was wondering if there was some sort of formula that can tell you how much wind speed is increased after it goes through a funnel. Also, does the length and size of the funnel affect anything? — Preceding unsigned comment added by 67.83.184.207 (talk) 17:15, 24 December 2014 (UTC)[reply]

If the material passing through the funnel was incompressible (as water is, nearly), the easy approach would be to realse that the same amount comes out as goes in in any time, so the speed would increase as the ratio of input area to exit area. With air, there will be some increase in air pressure, I think, so the increase in speed will not be as great as this.→86.157.199.240 (talk) 20:57, 24 December 2014 (UTC)[reply]

This is more a science question that a maths question. Pressure inside a constriction generally reduces as a necessary consequence of energy conservation. This is called Bernoulli's principle. —Quondum 12:53, 25 December 2014 (UTC)[reply]

As for the length and size, yes, those would have an affect. A longer funnel would mean a more gradual bend, so less of a change in direction to the air, therefore reduced friction, with the funnel slowing it down less. The ratio of the cross sectional areas of the two openings is obviously important, but so is absolute size, with more turbulence and friction caused by a slower funnel of the same proportions, due to a higher surface area to volume ratio of the air contained in the funnel at any one time.

And, in case it's not obvious, the air coming out of the funnel, if then unrestricted, will soon return to the initial pressure, volume, and wind speed, perhaps reduced a bit due to friction with the funnel.

And I should also mention that one common shape for funnels used for pouring liquids, two conic frustrums of different pitch, with a sharp angle between them [1], would generate a lot of turbulence, while a single conic frustrom, or perhaps a cylinder gradually blending into a conic frustrum, and that gradually blending back into a smaller cylinder, would be the way to reduce friction the most. This one has the cylinders at both ends, but lacks the gradual transitions: [2]. This one (ignoring the sphere), has gradual transition at the bottom, but not the top: [3]. StuRat (talk) 21:24, 24 December 2014 (UTC)[reply]

Thanks for the help, but is there a formula that says "If wind speed x enters a funnel with openings that have area y and z, then the wind speed will increase to ..."? — Preceding unsigned comment added by 67.83.184.207 (talk) 16:34, 26 December 2014 (UTC)[reply]

Sounds like they're ignoring all the subtlety of laminar versus turbulent flow, friction, and pressure changes, and just comparing the ratio of the opening sizes. In that case, they probably expect the wind velocity to increase in the same proportion that the area of the opening decreased. So, if the area was cut in half, the wind velocity would double. StuRat (talk) 06:35, 27 December 2014 (UTC)[reply]

What is the context of this question? If it was asked as a pure math question in an elementary class, then they would probably expect you to apply the simplifications mentioned by StuRat and say that the wind velocity increases by the ratio of the intake to outlet areas. ( V_out = V_in ⋅ A_in / A_out ) If, instead, you are looking for an accurate, real-world answer, then this is a problem in compressible fluid dynamics, and you should re-ask it on the Science Reference Desk. I generally don't like desk hopping suggestions, but in this case it would be appropriate and you might have better luck there. -- ToE 02:46, 28 December 2014 (UTC)[reply]

Calculating wind speed in the rain

Imagine that rain is falling exactly at a 45° angle. Does this mean that the wind is blowing at 32m/s, since its force is equal to that of gravity? This seems reasonable to me, but I just witnessed it falling slightly more horizontally than vertically, and the wind doesn't seem to be anywhere close to 115 km/hr. Nyttend (talk) 21:23, 24 December 2014 (UTC)[reply]

No, 32 m/s² is an acceleration, not a speed. If the air speed was equal to the rain's speed, that would give you a 45 degree angle, assuming the wind is horizontal and has had time to fully impart it's speed to the rain. Now just how fast raindrops fall (terminal velocity) would depend on many things, such as the drop sizes. StuRat (talk) 21:41, 24 December 2014 (UTC)[reply]

As Stu explains, it's irrelevant to this problem, but the acceleration due to gravity on the Earth isn't 32 m/s² anyway; it's actually 32 ft/s² or 9.8 m/s² (in each case rounding to 2 significant digits). — Preceding unsigned comment added by 65.94.50.4 (talk) 00:55, 25 December 2014 (UTC)[reply]

According to this table [4] from meteorologist Steve Horstmeyer, the maximum terminal velocity of raindrops is 9.09 m/s, or 20.2 mph. The terminal velocity depends on drop size, and as Horstmeyer explains here [5], the maximum drop size is 5 mm, as aerodynamic forces cause larger drops to break apart. AndyTheGrump (talk) 21:49, 24 December 2014 (UTC)[reply]

A good starting point, but seems like some false precision there. Certainly not all drops reach the same size. Temperature, humidity, wind, etc. would all play a part. Perhaps even if sunlight hits them might. StuRat (talk) 22:32, 24 December 2014 (UTC)[reply]