F
Fred Bloggs
- Jan 1, 1970
- 0
Larry said:My readings of the OP's description gave either 1000', 100' or 900'
of parallel pipe and power. My remarks apply to any of them. For
sake of discussion, suppose the power flows thru a pair of wires,
one located 11 inches from the pipe and the other 12 inches. (an
assumption corresponding to Romex with about 2 inch diameter)
For the moment, assume they maintain these distances along their
length. (no twisting) Assume a current +I at 11 inches, -I at 12.
The magnetic field in free space encircling one wire at radius r is:
B(r) = u0 * I / (2 pi r)
for an infinite straight wire where u0 = 1.26e-6 and r is in meters.
(Except near the ends of the run, this *closely* approximates the
situation given the assumed geometry.)
Evaluating for +I at 11/39.37 m and -I at 12/39.37 m, yields
B = (1.26e-6 * I / (2 * pi)) * (39.37 / 11 - 39.37 / 12)
B = I * 5.98e-8
Considering I to be the peak value of a 60 Hz AC current, and
so taking B(t) = sin(2 * pi * 60 * t) * I * 5.98e-8 , then
d(B(t))/dt = 2 * pi * 60 * cos(2 * pi * 60 * t) * I * 5.98e-8
= I * 2.25e-5 cos(...) Tesla/Second
This converts directly to Volts/Meter induced along the pipe.
(Doubters can start from F = q V x B, noting that the B field
changing in and out w.r.t. the wire pair is equivalent to the
V x B term. This is too basic to be spelled out here in detail.)
Taking the parallel run as somewhere between 30.5 and 305
meters, the induced peak voltage is 685 uV to 6.85 mV per
Amp of current flow.
With any twisting, that number will go down. If a twist were
applied to one end and the intermediate cable were follow in
proportion along its length, the coupling would be multiplied
by sinc(a), where a was the twist in radians.
Unless they are using thousands of Amps at that construction
site, the 30 VAC observed by the OP clearly has to be due
to some other effect or a markedly different geometry.
Some may opine or proclaim that only the integral form
of the induction law can be validly applied, or that twist
could be non-uniform, or the spacing may vary in that
trench. Except for the latter, none of it changes the
result because my use of the differential form already
assumes the worst possible return path (whereas the
real return path is coincident with the pipe, reducing
the net coupling between earth and pipe). Twisting
of any kind can only reduce the coupling; uniformity
simply causes more reduction, on average. If the OP
buried separate conductors willy-nilly with respect
to the pipe, that could be a problem. In Washington
state, that would never pass muster with the inspector
and I doubt local codes on this issue vary much.
If the OP's problem arose from mixing the power and
pipe badly, the result shows (in part) why that code
exists, and he has some digging to do. But I think it
much more likely that there is an exposed portion of
a hot power conductor leaking *a lot* of power into
ground and the pipe. That is an unstable situation,
dangerous as I've stated, and one the power company
will want fixed, if only because their meter is going to
be downstream from a ground-heating fault.
Given the OP's assertion that the 1000' pipe is buried
in red clay, I don't see any way for it to be anything
but a fault. Driving a well grounded conductor like
that to 25 VAC would take more power than even
an accidental air core transformer could deliver.
Typical horse-manure response that proves once again you can't even
understand what the OP wrote- and also that you love to shoot your mouth
off on subject matter about which you are clueless.