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RE: Old Popular Electronic Magazine
Original poster: Jim Lux <jimlux@xxxxxxxxxxxxx>
At 08:38 PM 2/19/2007, you wrote:
Original poster: "Jim Mora" <jmora@xxxxxxxxxxx>
Gary,
The high tension towers have 100's of mega watts (60hz, no relation to Tesla
coils) running through them that leads to a high electrostatic capacitive as
well as magnetic fields. I seem to remember stories of farmers winding coils
to be cut by this field. The losses would seem to be small compared to
corona losses.
However the magnetic field reduces by the inverse square of the distance
from the wire.
Actually, inverse of distance.. (it's a "infinitely long conductor"
carrying the current) so field goes as 1/r, not the 1/r^2 of a point source.
But a bigger problem is that the fields from the 2 or 3 conductors
cancel each other to a large degree (and the phases are rotated
through the positions on the towers to keep the C to ground balanced).
Typically, HV transmission lines have a characteristic impedance of
several hundred ohms, so if you know the voltage, you can guestimate
the current (for instance, the HVDC intertie terminating in Los
Angeles carries 3kA at 1MV, for an impedance of about 300
ohms). 1000 amps might not be a bad starting point for a HV transmission line.
Let's assume the utility is stupid, and runs a single wire carrying 1000 Amps
B= mu0* I/(2*pi*r)...
let's assume we're 50 feet (15 meters) from the wire...
B= 4*pi*1E-7 * 1000/(2*pi*15) or
= 2E-7*1000/15 = 133E-7 = 1.33E-5 Tesla (earth's mag field is about
0.5E-4 Tesla)
OK.. let's assume you've got a collecting loop with an area of 1
square meter. Faraday's law says that
EMF = N*dphi/dt where phi = B*A
B is actually 1.4cos(377t) * 1.33E-5
taking the derivative 1.4*377sin(377t)*1.33E-5 = 0.007 V/turn
OK.. now wind 100 turns, and make the area of the coil 100 square
meters... we're up to 70 Volts.. But what's the series resistance of
that coil.. Let's say we make the coil 10x10 meters, so each turn
uses 40 meters of wire, so we have about 4km of wire total in the
coil. Let's further assume we use AWG 10 copper wire, which has a
resistance of about 1ohm/1000ft.. 4km is about 13000 ft (3.28 ft/m)..
so, 13 ohms. If you short the coil, you'll get a current of 70/13
amps (5.4 A), dissipating about 400W in the coil.
And, of course, in reality, you'll be 15 meters from one conductor
and perhaps 20 meters from another conductor which will partially
cancel the field.