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Wind-powered Wi-Fi Sensors

Glenn Fleishman writes "According to an article at Indolink a 10-centimeter diameter windmill can produce the 7.5 milliwatts needed for a wireless sensor. The paper was published earlier (available as a PDF), but Nature magazine has apparently picked up the tidbit. The process flexes piezoelectric crystals to create a current. Although flywheels aren't mentioned in this article, it seems like a windmill, a flywheel, and a solar cell could in combination produce effective power in a range of conditions for remote wireless devices, including network relays obviating batteries entirely."

6 of 89 comments (clear)

  1. not enough power for 802.11 by j1m+5n0w · · Score: 4, Informative

    802.11 cards typically consume around 1 or 2 watts. They are probably targeting much simpler radios, like those used in motes.

  2. Re:Doing It The Hard Way? by EnderWigginsXenocide · · Score: 5, Informative

    Also found on the internet: " The piezoelectric generator is a much more efficient way of converting wind energy on a small scale than the conventional generators that create energy for the national power grid from wind turbines.

    A conventional generator that used a 10-cm turbine would convert only 1 per cent of the available wind energy directly into electricity. A piezoelectric generator ups that to 18 per cent, which is comparable to the average efficiency of the best large-scale windmills, says Priya. "

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  3. Re:electric double layer caps by Anonymous Coward · · Score: 2, Informative
    I'm trying to remember my electronics classes, but doesn't the capacitance drop in half if you put capacitors in series?


    You're correct, however: energy stored = 0.5CV^2 (C = capacitance, V = voltage), so if C decreases by a factor of 2 while V increases by a factor of 2 (making V^2 increase by a factor of 4), "energy stored" will overall increase by a factor of 2.
  4. Re:electric double layer caps by Anonymous Coward · · Score: 1, Informative

    The energy stored in a capacitor is

    U = Q^2/2C = C/2*V^2

    If you put your two identical capacitors in series, then each is dropping only half of the voltage of the battery (the same for each as the lower voltage before). But at the same time the capacitance is cut in half (by being in series). So the energy stored is 1/2 * (1/2C) * (V/2)^2 = 1/16 * VC. If you put your capacitors in parallel with half the voltage then the energy stored is 1/2 * 2C * (V/2)^2 = 1/4 * VC. In your example of 70 F and 2.1 V this is (bumping up the battery to 4.2 V for the series case) 77 J series vs 309 J parallel.

    Parallel wins for energy storage.

  5. Re:ummm by vespazzari · · Score: 2, Informative

    I cant seem to find the page but I was looking at flywheels a while back and a simple solution to the 2 problems you list where solved. Basically the flywheel was floating on magnets in a vacum in a steel drum. the company listed the life of the flywheel at 50 years with zero maintinance. basically you could bury them and that takes care of the vacation problem. I think i may have missed your point though, because i dont really understand why you think this would require a lightweight material or that we dont have it yet. it would be easy enough to create a really dense and strong flywheel and not push it to its limits, reducing the need to worry about the thing exploding.

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  6. Re:Larger applications? by odysen · · Score: 2, Informative

    Forget about wind turbines on a house. A house in a windy location will generally need about 1kw wind turbine. The turbine should on at least a 30 foot tower, above your roof. That being said, turbines on top of buildings usually aren't a good idea. The building becomes an acoustic from the vibrations of the wind turbine. Not to mention your home probably wasn't build to handle the stress. Anyway.. Solar is great solution though.