A voltage to ground mystery.

[Glenn Ashmore]

This is more of an electricity question than electronic but: We just
finished running water and power to a new construction site. Water is in a
copper pipe run about 1,000' through the woods burried about 2' deep that
just ends in a 2' vertical piece with a faucet on top. Power runs parallel
to the pipe about 100' away to a transformer on a pole at the construction
end.

The mystery is that there is a 25-30VAC difference between the faucet and
the mud puddle below it. Touch the faucet without wearing rubber boots and
you get shocked! I can see how a current might be induced into the pipe as
it runs along the power line but how can there be a potential between the
ends of a 24" copper pipe???

--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

 

[John]

"John Fields" <jfields@austininstruments.com> wrote in message
news:4fgg411ifukvgicc9vo54okqu4srsea79v@4ax.com...

On Mon, 28 Mar 2005 08:57:21 -0800, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:ls7g419rupl4j9ar8d5f9lbrjjnb8bjrop@4ax.com...
On Mon, 28 Mar 2005 09:42:21 -0500, "Glenn Ashmore" <gashmore@cox.net
wrote:

This is more of an electricity question than electronic but: We just
finished running water and power to a new construction site. Water is
in a
copper pipe run about 1,000' through the woods burried about 2' deep
that
just ends in a 2' vertical piece with a faucet on top. Power runs
parallel
to the pipe about 100' away to a transformer on a pole at the
construction
end.

The mystery is that there is a 25-30VAC difference between the faucet
and
the mud puddle below it. Touch the faucet without wearing rubber boots
and
you get shocked! I can see how a current might be induced into the
pipe as
it runs along the power line but how can there be a potential between
the
ends of a 24" copper pipe???

---
There isn't, there's a differential between the secondary of a
transformer (the 1000 feet of pipe) running in parallel with the
primary of the transformer (the power line) and the puddle.


There are two difficulties with that analysis.

1. The currents flowing in the power cable should be somewhat
well balanced and they flow within an inch or two of each other.
The conductors carrying those currents will have some degree
of twisting, tending to cancel when considering the net coupling
to any single parallel conductor. Magnetic coupling coefficients
sufficient to effect a 30/240 stepdown are highly improbable.

2. The copper pipe may be in the same trench, but it should be
separated by some distance. (I forget the code requirement,
but enforcing it is part of the reason an inspection is required
before the trench is filled in.) Unless the pipe is jacketed, (a
condition not stated by the OP), the fill around it will further
serve to reduce the coupling from nearby power conductors.

---
If it's not magnetic, then it's either capacitive crosstalk or
leakage.
---

As I stated in another post, the OP's observations could easily
be due to an unsafe condition that should be investigated by
a competent person on site, who can discover the additional
facts necessary to explain and correct the problem. This is
true regardless of any debate that might occur regarding the
magnetic coupling. I should hope that Mr. Fields is able to
concur with that view, even if he disagrees with my points
about magnetic coupling.

---
_Able_ to concur, you pompous ass? I'm able to do whatever I _choose_
to do, with or without any prompting from you. As a matter of fact,
I'd prefer not to be associated with you at all so, in the future,
stand on your own two feet and don't involve me in posts which require
rubber-stamping validation of your views. Glenn Ashmore isn't stupid,
and whether I _choose_ to concur with your view or not, I'm sure he'll
consider your view and do whatever's required to keep from killing
himself or someone else.


--
John Fields

Right on!

 

[Glenn Ashmore]

Well this is in the Piedmont of Georgia so the pipe is definitely burried in
1,000' of red clay.

The thing I can't figure out is the copper pipe comes out of the ground
right at the puddle so it should be well grounded but if you use a VOM
between the faucet and a nail stuck into the mud a foot from the base it
reads 25V!

--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

 

[Larry Brasfield]

"Glenn Ashmore" <gashmore@cox.net> wrote in message
news:gJU1e.80145$SF.30120@lakeread08...

This is more of an electricity question than electronic but: We just
finished running water and power to a new construction site. Water is in a
copper pipe run about 1,000' through the woods burried about 2' deep that
just ends in a 2' vertical piece with a faucet on top. Power runs parallel
to the pipe about 100' away to a transformer on a pole at the construction
end.

The mystery is that there is a 25-30VAC difference between the faucet and
the mud puddle below it. Touch the faucet without wearing rubber boots and
you get shocked! I can see how a current might be induced into the pipe as
it runs along the power line but how can there be a potential between the
ends of a 24" copper pipe???

To determine what is going on from a distance is difficult,
without knowing a few more specifics, so maybe you could
answer a few questions to help narrow it down.

1. Does the voltage you measure between that faucet and
"earth" (the puddle, for instance) vary significantly as you
add or remove load from your construction pole? The no
load case is best for distinguishing magnetic coupling, being
pretty much ruled out it the 25-30 VAC occurs without any
load at the far end of that power run.

2. Does that voltage vary if you load it, for example with an
low wattage incandescant lamp? (You would connect the
lamp between the faucet and a good local ground. Under
ordinary circumstances, that faucet should be a good ground,
but you will have to find another.

3. I presume there is a power pole at your site, and that you
have a small breaker panel there, and that there is a staked-
into-earth ground connection right there. Is that true? If not,
what is at the site end of that 1000' power line?

Some people here have implied that what you see is normal
or harmless. In the face of uncertainty about what is really
going on, and not knowing whether the coupling mechanism
could get better or not, I believe prudence calls for leaving
that exposed faucet alone, perhaps even covering it, until a
solution or convincing and harmless explanation is found.

By the way, my hope regarding concurance, (in my 8:57
post, poorly received by Mr. Fields) is about avoiding
needless confusion about the hazard you face. Of course
you can and will make your own decisions, but I assume
your question appeared here because you felt you could
use or needed some technical advice. If that means you
must choose and pick what to believe without knowing
for sure, I urge you to err toward safety.

What is most odd about your situation is that buried
conductors in contact with earth are often considered
to be good ground connections. Seeing 30 VAC on
a conductor with 1000' of earth contact means either
a very dry, high resistance connection or something is
driving it with some ability to drive power into it. If
that something is an errant and now poor connection,
it could become a better connection in the presence
of water or after some motion, say a truck running
over the trench where the illicit short is. If there is
such a connection, able to deliver power to an earthed
copper rod, it takes little imagination to to see either it
getting better or the earthed rod getting more poorly
connected so as to increase the AC voltage. It takes
only 5 to 10 mA in the right place to upset a normal
human heart rhythm, (and less for some people), so the
situation you describe is one dangerous enough that I
would not tolerate it if anybody I cared about was to
exposed to the hazard.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Roger Johansson]

"Glenn Ashmore" <gashmore@cox.net> wrote:

Well this is in the Piedmont of Georgia so the pipe is definitely
burried in 1,000' of red clay.

And the copper pipe is not isolated from the clay in any way?
Is it covered with some kind of laquer which may be invisible?

The thing I can't figure out is the copper pipe comes out of the ground
right at the puddle so it should be well grounded but if you use a VOM
between the faucet and a nail stuck into the mud a foot from the base
it reads 25V!

You could do some experiments to find out more about this problem.

Disconnect the cable from the mains so it carries no voltage anymore,
then measure again. If the voltage on the pipe disappears we know that
the voltage comes from the cable somehow.

Try connecting a 1k resistor between the pipe and the nail in the ground,
then measure the voltage again. If it is still 25V you have a dangerous
pipe there, capable of delivering current as well as voltage.

Check up the possibility that the pipe is making contact to some mains
wiring under ground or before it goes into the ground. The voltage must
come from somewhere. And it is very unlikely that it is done through
induction only, for reasons others have explained.



--
Roger J.

 

[John Popelish]

Glenn Ashmore wrote:
(snip)

The thing I can't figure out is the copper pipe comes out of the ground
right at the puddle so it should be well grounded but if you use a VOM
between the faucet and a nail stuck into the mud a foot from the base it
reads 25V!

But that reading tells you nothing about the isolation between the
pipe and the point a foot away. By ohm's law, if there were a
milliohm if resistance between the two points, that voltage would
imply 25 amperes passing between them. If the resistance were 1 meg
ohm, that voltage would imply 25 microamperes between them. The real
resistance is almost certainly between these limits, so the current is
somewhere between those values. Your problem is not only to find a
source of 25 volts AC, but a source that can support a significant
current load.

In the mean time, you may be able to lessen the shock symptoms while
you search for the cause by placing a few short ground rods around the
puddle, and tie them all to the pipe. This will increase the load
current on the pipe, but it will lower the drop across that particular
segment of the circuit.

--
John Popelish

 

[Chris]

"Glenn Ashmore" <gashmore@cox.net> wrote in message
news:gJU1e.80145$SF.30120@lakeread08...

This is more of an electricity question than electronic but: We just
finished running water and power to a new construction site. Water is in
a
copper pipe run about 1,000' through the woods burried about 2' deep that
just ends in a 2' vertical piece with a faucet on top. Power runs
parallel
to the pipe about 100' away to a transformer on a pole at the construction
end.

The mystery is that there is a 25-30VAC difference between the faucet and
the mud puddle below it. Touch the faucet without wearing rubber boots
and
you get shocked! I can see how a current might be induced into the pipe
as
it runs along the power line but how can there be a potential between the
ends of a 24" copper pipe???

--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com


Another idea.

Have you tried to connect a wire between the faucet and the base of the
pipe?

You may have plumbers tape insulating the faucet from the pipe.

Cheers.

 

[Glenn Ashmore]

"John Popelish" <jpopelish@rica.net> wrote in message
news:4248B432.ADD76045@rica.net...

But that reading tells you nothing about the isolation between the
pipe and the point a foot away. By ohm's law, if there were a
milliohm if resistance between the two points, that voltage would
imply 25 amperes passing between them. If the resistance were 1 meg
ohm, that voltage would imply 25 microamperes between them. The real
resistance is almost certainly between these limits, so the current is
somewhere between those values. Your problem is not only to find a
source of 25 volts AC, but a source that can support a significant
current load.

Kinda frightning when you do the math! That could kill! There can't be
more than a couple of milliohms resistance between the ground and the
faucet. That means there must be a lot of current.

The pipe is brand new 1 1/4" soft copper tube running from the meter. I
burried it myself with a Ditchwitch and put a tee with a 1/2" temporary
stand pipe at the end for use during construction. All the joints are
soldered. No problem until the power company came in last week to plant
their poles and hang line. The line is about 50' to one side and the
transformer pole is back about 100'. Single phase service so there is one
insulated wire and a bare cable to the transformer.

I can see how that arangement can induce a current in the pipe but I would
think it would bleed off into the ground. I will call the power company
tomorrow. It does not sound like anything I want to fool with.

--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

 

[Steve Dunbar]

Glenn Ashmore wrote:

....

The mystery is that there is a 25-30VAC difference between the faucet and
the mud puddle below it. Touch the faucet without wearing rubber boots
and you get shocked! ...
...

Ouch! Try doing a Google search for "stray voltage."

--
-- Steve

 

[Rich Grise]

On Tue, 29 Mar 2005 00:12:50 -0500, Glenn Ashmore wrote:

I can see how that arangement can induce a current in the pipe but I
would think it would bleed off into the ground. I will call the power
company tomorrow. It does not sound like anything I want to fool with.

Yes, this is the best suggestion yet.

I was getting my brain all in a bunch here: "What's the resistivity
of "Good ol' Georgia Red Clay" and stuff -

But "check with somebody who knows what they're talking about" is
probably the wisest option in this case.

Good Luck!
Rich

 

[John Popelish]

Glenn Ashmore wrote:

"John Popelish" <jpopelish@rica.net> wrote in message
news:4248B432.ADD76045@rica.net...

But that reading tells you nothing about the isolation between the
pipe and the point a foot away. By ohm's law, if there were a
milliohm if resistance between the two points, that voltage would
imply 25 amperes passing between them. If the resistance were 1 meg
ohm, that voltage would imply 25 microamperes between them. The real
resistance is almost certainly between these limits, so the current is
somewhere between those values. Your problem is not only to find a
source of 25 volts AC, but a source that can support a significant
current load.

Kinda frightning when you do the math! That could kill! There can't be
more than a couple of milliohms resistance between the ground and the
faucet. That means there must be a lot of current.

A couple milliohms between that bit of ground and the faucet? What
makes you say that? I have worked very hard with a couple thousand
feet of buried copper to get the ground resistance (whole Earth to
copper resistance, not resistance from some small patch of ground to
copper) below 5 ohms.

The pipe is brand new 1 1/4" soft copper tube running from the meter. I
burried it myself with a Ditchwitch and put a tee with a 1/2" temporary
stand pipe at the end for use during construction. All the joints are
soldered. No problem until the power company came in last week to plant
their poles and hang line. The line is about 50' to one side and the
transformer pole is back about 100'. Single phase service so there is one
insulated wire and a bare cable to the transformer.

Are you saying that this service buried a bare conductor and an
insulated conductor along side the pipe? If the service has a high
leakage current from the primary side to the supposedly isolated
service, it may be dumping quite a bit of current into the Earth
through that bare neutral wire. The power company should check that
out at the pole, to determine how much AC current it is passing to the
neutral with no load connected at your end. This may well be the
problem. And if it is, they would want ot know about it, because it
is an unmetered power consumer they are carrying.

I can see how that arangement can induce a current in the pipe but I would
think it would bleed off into the ground. I will call the power company
tomorrow. It does not sound like anything I want to fool with.

Agreed. It ain't normal.

--
John Popelish

 

[Roger Johansson]

John Popelish <jpopelish@rica.net> wrote:

But that reading tells you nothing about the isolation between the
pipe and the point a foot away. By ohm's law, if there were a
milliohm if resistance between the two points, that voltage would
imply 25 amperes passing between them.

You meant ohm instead of milliohm here, didn't you?
I saw it before but didn't say anything because it had a positive effect,
he realized how dangerous it is. (25V=25A*1Ohm)

Kinda frightning when you do the math! That could kill! There can't
be more than a couple of milliohms resistance between the ground and
the faucet. That means there must be a lot of current.

A couple milliohms between that bit of ground and the faucet? What
makes you say that?

Maybe you made him think and say that yourself, see above.


--
Roger J.

 

[Fred Bloggs]

Glenn Ashmore wrote:

"John Popelish" <jpopelish@rica.net> wrote in message
news:4248B432.ADD76045@rica.net...


But that reading tells you nothing about the isolation between the
pipe and the point a foot away. By ohm's law, if there were a
milliohm if resistance between the two points, that voltage would
imply 25 amperes passing between them. If the resistance were 1 meg
ohm, that voltage would imply 25 microamperes between them. The real
resistance is almost certainly between these limits, so the current is
somewhere between those values. Your problem is not only to find a
source of 25 volts AC, but a source that can support a significant
current load.


Kinda frightning when you do the math! That could kill! There can't be
more than a couple of milliohms resistance between the ground and the
faucet. That means there must be a lot of current.

The pipe is brand new 1 1/4" soft copper tube running from the meter. I
burried it myself with a Ditchwitch and put a tee with a 1/2" temporary
stand pipe at the end for use during construction. All the joints are
soldered. No problem until the power company came in last week to plant
their poles and hang line. The line is about 50' to one side and the
transformer pole is back about 100'. Single phase service so there is one
insulated wire and a bare cable to the transformer.

I can see how that arangement can induce a current in the pipe but I would
think it would bleed off into the ground. I will call the power company
tomorrow. It does not sound like anything I want to fool with.

The power grid is circulating current through the GND returns at the
two different points of connection to the grid- one at the origin of the
copper pipe and the other at the construction site. You have created a
nice short-circuit with that pipe and the circulating current can be
very large. You will have no option except to install a section of
non-conducting plastic pipe at the construction site, make the faucet
and its T-section riser plastic also.

 

[Larry Brasfield]

"Rich Grise" <richgrise@example.net> wrote in message
newsan.2001.07.12.14.58.03.396981@example.net...

On Mon, 28 Mar 2005 08:57:21 -0800, Larry Brasfield wrote:
"John Fields" <jfields@austininstruments.com> wrote in message
On Mon, 28 Mar 2005 09:42:21 -0500, "Glenn Ashmore" <gashmore@cox.net

This is more of an electricity question than electronic but: We just
finished running water and power to a new construction site. Water is in a
copper pipe run about 1,000' through the woods burried about 2' deep that
just ends in a 2' vertical piece with a faucet on top. Power runs parallel
to the pipe about 100' away to a transformer on a pole at the construction
end.
---
There isn't, there's a differential between the secondary of a
transformer (the 1000 feet of pipe) running in parallel with the
primary of the transformer (the power line) and the puddle.

There are two difficulties with that analysis.

1. The currents flowing in the power cable should be somewhat
well balanced and they flow within an inch or two of each other.
The conductors carrying those currents will have some degree
of twisting, tending to cancel when considering the net coupling
to any single parallel conductor. Magnetic coupling coefficients
sufficient to effect a 30/240 stepdown are highly improbable.

Don't forget the 900' of pipe running parallel to the hiline upstream
of the transformer.

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.

Cheers!

You bet!

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Fred Bloggs]

Larry Brasfield wrote:

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.

 

[Larry Brasfield]

"Fred Bloggs" <nospam@nospam.com> wrote in
message news:42492432.7010007@nospam.com...

Larry Brasfield wrote:

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.

I note that Mr. Bloggs appears unable to point out any
specific fact I have misunderstood from the OP's posts,
nor has he specified any particular flaw in my analysis.
He is fond of emitting noise without substance, mainly
for love of viewing his own spew, I think.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Fred Bloggs]

Larry Brasfield wrote:

"Fred Bloggs" <nospam@nospam.com> wrote in
message news:42492432.7010007@nospam.com...

Larry Brasfield wrote:


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.



I note that Mr. Bloggs appears unable to point out any
specific fact I have misunderstood from the OP's posts,
nor has he specified any particular flaw in my analysis.
He is fond of emitting noise without substance, mainly
for love of viewing his own spew, I think.

Maybe in your world- but as usual- you're a worthless crock of excrement
when judged against reality- and your posts convey nothing of any use or
value. It's a sorry-sack of manure who reaches nearly 60 yrs of age and
still fails at communication as badly as you do. But you don't worry
about that- just keep pecking away at your sickening and pompous
garbage- it's all about some fantasy you entertain where people just
fall over admiring your pretense of expertise..

 

[Glenn Ashmore]

"John Popelish" <jpopelish@rica.net> wrote in message
news:4248FE1E.35C5E032@rica.net...

<snip>

Are you saying that this service buried a bare conductor and an
insulated conductor along side the pipe? If the service has a high
leakage current from the primary side to the supposedly isolated
service, it may be dumping quite a bit of current into the Earth
through that bare neutral wire. The power company should check that
out at the pole, to determine how much AC current it is passing to the
neutral with no load connected at your end. This may well be the
problem. And if it is, they would want ot know about it, because it
is an unmetered power consumer they are carrying.

The power line is up about 30' on poles.

Just called the power company. When they figure it out I will report back.

--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

 

[John Popelish]

Roger Johansson wrote:

John Popelish <jpopelish@rica.net> wrote:

But that reading tells you nothing about the isolation between the
pipe and the point a foot away. By ohm's law, if there were a
milliohm if resistance between the two points, that voltage would
imply 25 amperes passing between them.

You meant ohm instead of milliohm here, didn't you?
(snip)

I will blame the thousand to one error on sitting here dissolving with
the flue.
But I think the point about the voltage not telling him what the
ground current is made an impression.

--
John Popelish

 

[Roger Johansson]

"Glenn Ashmore" <gashmore@cox.net> wrote:

The power line is up about 30' on poles.

There has been a lot of misunderstanding in this thread, and I am not
sure where it started. I think you started the confusion yourself, by
talking about induction. That would only be remotely possible if the
conductors were very close to each other and buried together.

The only important part of the thread was that you were made aware of how
serious the situation is and that you need get it fixed by competent
people.


--
Roger J.