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# Wind heater

Funnely.
 (-1) [vote for, against]

Prompted by the linked idea, which suggests using a small wind-turbine to generate electricity (and thence heat) from a chilling arctic wind.

There is a better way.

Imagine a large funnel, consisting of a large hoop with a plastic windsocky thing attached to it. The small end of the funnel is connected to a pipe. The pipe is closed at the other end, apart from a small hole.

When facing into the wind, this arrangement will compress the air in the pipe; the stronger the wind, the greater the compression.

The (absolute) temperature of a gas rises as its compressed. If the funnel raises the pressure of air going through the pipe from 15psi to 18psi, then the temperature of the air in that pipe will increase by a factor of 18/15. Thus, air at -20°C (253°K) will be heated to +30°C (303°K).

This must surely be the simplest way of extracting heat Just run the hot pipe through your tent (or even through your sleeping bag) and you're toasty warm even in the arcticest of winds.

 — MaxwellBuchanan, Nov 03 2012

[MaxwellBuchanan, Nov 03 2012]

Wind Powered Fridge (no moving parts) Wind_20Powered_20Fr...o_20moving_20parts)
Could also serve as a heater [csea, Nov 04 2012]

And what precisely would be the proposed dimensions of such an unlikely contraption?
 — Alterother, Nov 03 2012

math for windspeed for +3psi from a funnel ?
 — FlyingToaster, Nov 03 2012

 //what precisely would be the proposed dimensions//

Mouth: 2m diameter. Funnel length: 4m. Funnel exit diameter, and pipe internal diameter: 2cm. Exit hole from pipe: 4.5mm.
 — MaxwellBuchanan, Nov 03 2012

Aaagh!
 — spidermother, Nov 03 2012

See, I told you not to put your hand on the pipe.
 — MaxwellBuchanan, Nov 03 2012

 //what precisely would be the proposed dimensions// OK, seems reasonable.

 //math for windspeed for +3psi from a funnel ?// Yes, please, what windspeed?

I think we went through this with a cooling intent (though not in a tent!) using a Hilsch Vortex device. [link]
 — csea, Nov 04 2012

 OK, I admit that the maths is a little outside my scope.

 However, a windspeed of 20mph is stated to give a wind *pressure* of around 100 Newtons per square metre, or about 0.02psi.

 We are aiming for a pressure rise of 3psi, or 150 times greater than the wind pressure. At first glance, this suggests that a 150-fold reduction in area would achieve this. A 2m funnel feeding into a 4.5mm exit hole would give an area reduction of over 200,000 - much much more than 150-fold.

 But of course it's not nearly that simple, and the 200,000 fold is a red goose.

 So the question is, how do you determine the pressure at the exit point of a funnel, given the dimensions of the funnel and the speed of the impinging wind?

 There must be *some* pressure increase at the exit of the funnel, even though it will not be proportional to the area reduction.

As a final argument, bear in mind that there is at least as much energy available from the wind as we could extract using a wind turbine. Since a 2m wind turbine in a strong wind will generate a useful amount of power, it follows that this idea is not energetically forbidden.
 — MaxwellBuchanan, Nov 04 2012

 Bother. I can't tell whether you're just trolling now, but I'll trip trap over that bridge when I come to it.

Anyway, 20mph at 293K gives a stagnation pressure of about 48 Newtons per square metre (above the static air pressure). Someone forgot to divide by 2. And that's the absolute maximum pressure attainable - the narrower the constriction, the closer you get to that value.
 — spidermother, Nov 04 2012

 Any simple tube/funnel/Nutrimatic cup facing into the wind at steady state produces pressure at most œ density * v² above the static pressure. Allowing any flow (which is necessary to harvest energy) reduces the pressure.

 Of course it's possible to produce greater pressures, but you need a mechanical advantage - you need to pressurise a small flow of air by harvesting power from a large flow of air, for example by linking a compressor to a turbine. A steady-state funnel can't do that, as the two flows are the same.

It is also theoretically possible to use resonance, as a hydraulic ram pump does. But however you do it, there will be a high-volume, large area, low pressure waste flow as well as a low-volume, high pressure flow. There is no way around that.
 — spidermother, Nov 05 2012

If you were to use the wind to compress the air by some mechanical means, such as a belt-driven piston-type compressor, you'd get heat off the outside of the cylinder, or just the outlet pipe. You could use that heat, as in the original idea, then you'd have a nice tank of compressed air for blowing dust off your keyboard.
 — baconbrain, Nov 05 2012

Good point, [bacon]. Plus, if you extracted the heat from the compressed air and then allowed it to re- expand, you could have a freezer for your wodka.
 — MaxwellBuchanan, Nov 05 2012

My wod is already pretty chilly at this time of year, ka.
 — pocmloc, Nov 05 2012

 //Any simple tube/funnel/Nutrimatic cup facing into the wind at steady state produces pressure at most œ density * v² above the static pressure.//

 I've come back to this because the above equation is pertinent to some other problems I'm thinking about. So, if I have a wind speed of 5m/s, and if air density is roughly 1kg/m3, I can get a pressure of œ * 1 * 5² = 12 N/m² above ambient? Or have I got a unit wrong somewhere?

And since the shape of the 'funnel' doesn't enter into the equation, doesn't that mean that the pressure at the bottom of a 1m² funnel is the same as the pressure on the face of a flat board?
 — MaxwellBuchanan, Jul 09 2013

AFAIK the *average* pressure on the board will be closeish to the tube/funnel pressure. But assuming no work is done, the pressure near the edge of the board would be lower, since the air is moving faster.
 — spidermother, Jul 10 2013

 OK, on further reading, is seems that the pressure at the centre of the plate is equal to the stagnation pressure, but the average pressure is somewhat less (because the pressure decreases towards the edges, as stated above).

 Note that this applies to isentropic (no work is done), incompressible flow - which is closely approximated for static objects and flows well below the speed of sound.

And yes, your calculation appears to be correct.
 — spidermother, Jul 10 2013

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