Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Path: utzoo!mnetor!uunet!lll-winken!lll-lcc!ames!sunybcs!kitty!larry From: larry@kitty.UUCP (Larry Lippman) Newsgroups: sci.electronics Subject: Re: Living near high tension lines Message-ID: <2209@kitty.UUCP> Date: Sat, 7-Nov-87 15:37:26 EST Article-I.D.: kitty.2209 Posted: Sat Nov 7 15:37:26 1987 Date-Received: Mon, 9-Nov-87 07:14:22 EST References: <9312@tekecs.TEK.COM> <1718@bloom-beacon.MIT.EDU> <1913@frog.UUCP> Organization: Recognition Research Corp., Clarence, NY Lines: 84 Summary: "Free Electric Power" - not likely... In article <1913@frog.UUCP>, die@frog.UUCP (Dave Emery) writes: > >you could do what a buddy of mine did... > >(his barn was overshot by high voltage lines) > >wrap the rafters with coils, and make off with some free power!! > >his barn had "infinite" lighting this way!!! One minor point: no way, EVER will an electric utility company run power lines OVER a customer structure (the liability factors and other reasons for this should be obvious). > I have heard a similar story about farmers in the Midwest using > long stretches of barbed wire fence near long distance power lines for the > same "free" power gambit. As I understand it the power companies were able > to sucessfully prosecute some of them for stealing power. The above situations sound like rural (as opposed to urban) legends to me. I suppose if I keep this up, someone is going to confer upon me the title of "Net Skeptic"... :-) [This remark, which I couldn't help, makes more sense if you have been following my postings in another newsgroup, which shall remain unnamed :-).] Now, let's look at this situation and apply a bit of electrical engineering. Let's assume the following: 1. There is a high-voltage electrical transmission line operating at 128 to 356 kV; the actual voltage is immaterial since all we care about is current flow. A typical single conductor current for such a transmission line using 1200 MCM ACSR conductor is 1,000 amperes (somewhat on the high side). 2. There is a distance of 40 feet from the power conductor to a "receiving coil"; this figure is pretty conservative, since under most circumstances, I don't believe it is possible to get that close. Since this amounts to a one-turn primary, the flux density (B) 40 feet away will be given by: B = (4.52)(I)/(h), where B is flux density in lines/sq inch, I is current in amperes, and (h) is "air gap" distance in inches. Plugging in 1,000 amperes and 40 feet makes B = 9.4 lines/sq in (1.5 Gauss for you Gauss fans). As a point of comparison, this magnetic field is roughly 3 times that of the earth's natural magnetic field. Now, let's make the following assumptions about a "receiving coil": 1. The coil is rectangular, 40 feet long and 4 feet wide, and optimally oriented below the transmission line conductor. 2. There is an iron "core" (pretty damn big piece of metal!); the resultant value below will be more than for an air core. Let's see what open-circuit sinusoidal voltage will appear in ONE turn of the above "receiving coil". E = (4.44x10^-8)(a)(B)(f)(N), where E is induced voltage, a is cross-sectional area of core, B is peak flux density in lines/sq in, f is frequency in Hertz, and N is number of turns (1 for now). Plugging in 9.4 lines/sq in of flux density and the area of the example "receiving coil" gives us an induced voltage of 0.58 volts per turn of receiving coil. Since 1/2 volt won't be very useful, we will need 120/0.58 = 206 turns to give us 120 volts. Let's use # 12 AWG wire. (206 turns)(40 ft + 40 ft + 4 ft + 4 ft) = 18,128 feet. Hmmm... That's OVER THREE MILES of wire to give us 120 volts. Let's see how much power we can derive: 18,128 feet of # 12 AWG wire gives us a DC resistance of 30 ohms, which means that by virtue of DC resistance alone, we will be limited to a power transfer of roughly 240 watts (assuming load impedance = source impedance). The power transfer will really be less, but I don't feel like doing the AC-side calculations to provide an exact figure. The above assumes an iron core in the "receiving coil", which at 4 x 40 feet in dimensions is pretty impracticable. Without actually doing the calculations, an air core would be at least an order of magnitude less efficient. Of course, we do have THREE conductors, each of which contributes to the magnetic field. On the other hand, no way will there be a 40 foot air gap between each conductor and the receiving coil. At BEST, with over 3 miles of # 12 AWG wire one may get enough free energy to light a 40-watt light bulb. Hardly seems worth the effort... At a cost of around $ 1,000.00 for the copper wire alone, I doubt that anyone has really tried it. [P.S. to Ed Hall: I will not be offended if you view this article with "suspicion" and doubt.] <> Larry Lippman @ Recognition Research Corp., Clarence, New York <> UUCP: {allegra|ames|boulder|decvax|rutgers|watmath}!sunybcs!kitty!larry <> VOICE: 716/688-1231 {hplabs|ihnp4|mtune|seismo|utzoo}!/ <> FAX: 716/741-9635 {G1,G2,G3 modes} "Have you hugged your cat tor,, but