Is the Earth a Resistor or Capacitor for Underground Cable and Pipe Location?

[mcstatz5829]

I work for a company that locates underground cables and pipes. We apply an AC signal to the underground facility and to be able to detect the location.

Take the simplified drawing above. Two transformers grounded with a buried electrical cable. We use a transmitter and connect it to one transformer. There's clearly a current divider here - some of the current will go out the transformer ground and some will carry on the metal sheath of the electrical cable to the far transformer.

I have an argument with some of my colleagues about the concept of ground path return. They say the electrical signal must return to the transmitter ground to complete the circuit. To me that is modelling the Earth as a resistor. That would certainly be true if we used a DC signal.

But we use an AC signal, typically 8 - 30 kHz, but sometimes as low as 500 Hz. I argue the Earth is better modeled as a capacitor, and the signal does not necessarily travel back to the ground stake.

What say you?

 

[anorlunda]

It is true that the capacitance to ground of underground power cables is big. In fact, that is often the limiting factor that determines how much power we can transmit how far underground. That is at 50-60 hz.

This is the equivalent circuit we use for power lines, both overhead and underground. Put N of those equivalents in series, and you have a distributed model of the line.

 

[Merlin3189]

However you look at the earth, your generator produces a signal that drives antiphase currents of equal magnitude out of its two terminals. So it seems odd to me to say the signal doesn't come back to the ground terminal. I might even say, the signal goes out from both terminals simultaneously and maybe never comes back.

Now, whether you model the circuit between the output terminals as a capacitor or a resistor, is another question.
A good Earth connection at both ends should appear as a low resistance between the output terminals. My understanding is that all the significant resistance is in and around the Earth connection. Beyond that there are so many parallel routes that, however resistive, the net resistance is effectively zero. (I see this as analagous to the field of a solenoid only being significant inside and near to the ends. Beyond that the field is spread over such a large area, that the flux density is effectively zero.)

So I would be inclined to model your cable like this:

where the wire resistors and parallel capacitors are many, expressed by the resistance and capacitance per unit length of the cable.

PS. I see Anorlunda has replied showing the series inductance, which I'd forgotten about.

 

[mcstatz5829]

So question I think is whether A) or B) is more correct.

I think as F increases, Rearth decreases, and Cearth increases - the models converge.

My interpretation is model B - there is no return through the ground. The Earth is simply absorbing and supplying charge like a capacitor.

But these guys may disagree with me: https://en.wikipedia.org/wiki/Single-wire_earth_return

However you look at the earth, your generator produces a signal that drives antiphase currents of equal magnitude out of its two terminals. So it seems odd to me to say the signal doesn't come back to the ground terminal. I might even say, the signal goes out from both terminals simultaneously and maybe never comes back.

Yes, that's exactly how I think of it. The transmitter is emitting two waves into the Earth with a 180 phase angle. Might come full circle... depending on soil conditions, distance, presence of other sources of high or low potential... might not...

 

[anorlunda]

Those schematics are misleading without putting numbers on the components.

The underground conductor is very close to the ground; say one inch. Those are the capacitors I showed in my diagram.

The return path, which you show as R_earth or C_earth represent long distances, perhaps miles. So the impedance is very high.

there is no little return through the ground. The Earth is simply absorbing and supplying charge like a capacitor.

I endorse that. That makes the sketches A and B unnecessary.

By the way, there will be a resonant frequency caused by the series L and ground C in the cable. It that what you guys do, vary the frequency looking for the max signal?

 

[mcstatz5829]

By the way, there will be a resonant frequency caused by the series L and ground C in the cable. It that what you guys do, vary the frequency looking for the max signal?

Resonant frequency - possibly, I would have to ask the equipment manufacturers if they use that for some of the detection.
We don't modify the frequency. The receiver has multiple antennas for detecting the magnetic field.

Simple version: The user moves it until it's registered a peak response, and that's usually where the facility is.

More complex: If there is capacitive coupling with another facility or they are commonly grounded, the magnetic field will get distorted and that affects the accuracy.

 

[anorlunda]

If there is capacitive coupling with another facility

I'm not sure what you mean by another facility. Do you mean another cable?

 

[jim hardy]

I have an argument with some of my colleagues about the concept of ground path return. They say the electrical signal must return to the transmitter ground to complete the circuit. To me that is modelling the Earth as a resistor. That would certainly be true if we used a DC signal.

By "signal" do you mean current ?
Seems to me Kirchoff dictates that current must get back to whence it came.

Put yourself inside the signal generator.

Any charge exiting to the right must be replaced by one entering from the left...
irrespective of what it flows through to get theredid i misunderstand the question ?

old jim

 

[anorlunda]

Seems to me Kirchoff dictates that current must get back to whence it came.

I think this is one of those rare exceptions where we can't use KCL or KVL. It is not a closed circuit. In order for this application to work, there must be energy (in the form of a tone) leaked from the cable to an above-ground detector along the whole run.

We are using the Earth as a source/sink for small charges for a short time. The only analogy I can think of is a Van de Graff generator where we (unidirectionally) pump charge to the dead end of an open circuit.

 

[jim hardy]

The only analogy I can think of is a Van de Graff generator where we (unidirectionally) pump charge to the dead end of an open circuit.

Lightning is similar. Charge accumulates someplace then returns all at once.
I told my students "..Kirchoff Rules!. But not necessarily immediately - he will take a rain-check. "

 

[anorlunda]

Jim, you just reminded me of something else we can tell students. The very definition of the word circuit includes closed. So open circuit is an oxymoron and closed circuit is redundant. We could use that to remind students that you can only apply circuit analysis on closed paths.

 

[Baluncore]

I argue the Earth is better modeled as a capacitor, and the signal does not necessarily travel back to the ground stake.

The signal always flows through the entirety of all of the available circuits. The buried UG conductor being followed concentrates the current and feeds the many return paths back to the Earth reference stake at the generator. The portable detector probably senses the magnitude and orientation of the magnetic field generated locally by current flowing in the nearby underground conductor, current that is on it's way to the many return paths beyond that point, or the earthed far end of the line.

It does not matter how you model the Earth return circuits as the detection is not phase sensitive. You are only interested in the magnitude of the AC current and the orientation of the local AC magnetic field due to UG conductor position.

If the cable is insulated from the earth, the impedance of the many Earth return paths will be made through the capacitance of the insulation in series with the Earth resistance. The generator will also see the cable inductance in series with that. The series LCR return path will have a lowest impedance at the frequency when the positive reactance of L and the negative reactance of C sum to zero. That will drive the maximum AC current through the UG conductor, to provide the best signal.

 

[anorlunda]

@Baluncore might be right, and I might be 100% wrong.

I'm thinking of a single system able to trace healthy, shorted, and sheared underground cables.

 

[Merlin3189]

I hope my comment,

I might even say, the signal goes out from both terminals simultaneously and maybe never comes back.

does not lead OP to think that I don't see this as a closed circuit.
What came into my mind to make me add that final clause was, if you send a signal down a transmission line into a matched load, then there is no reflection. The signal has gone and not returned. Current flowed equally and oppositely in both terminals at the start of the line and similarly at the end of the line.

Back to post #4, and the equivalent circuit models:
I prefer the Earth as a resistance, concentrated at the stakes.
Since the capacitance is between the cable and the immediate Earth surrounding, I'd agree with some local resistances, but more as shown in my diagram. I still see the bulk of the Earth as a low resistance.

One reason I can't see the Earth as a series of capacitors is the distance. If the capacitors bridge significant gaps, their capacitance is going to be very small - C∝1/d Then there will be thousands of small capcitances in series, making an even smaller total capacitance. If they bridge smaller distances, there will be proportionately more in series and the net capacitance will still be tiny.
As above, of course i see capacitance between the wire and the ground, but that is v.diff from your B cct.

 

[anorlunda]

I still see the bulk of the Earth as a low resistance.

I'm thinking of 1-10 km cables in soil. Soil resistivity about 1000 ohms/meter. Cable resistivity about 0.1 ohms/meter. Soil to cable resistance ratio about 10⁴

 

[Nik_2213]

The Earth's distributed electrical behaviour is complex but, IMHO, several replies have described its nuances very well.

I came across some of the quirks in a popular electronic magazine's series on 'earth phones', starting with sundry telegraphic experiments for communicating across canals or with islands via induction, then exploring the systems the English and French used during WW1 to communicate with their forward trenches. And, yes, it discussed the bloody debacles after the Germans, who deployed similar systems, heard whispers of cross-talk, added amplifiers and eaves-dropped extensively...

Modern applications include one type of 'cave phone', where a team just emplaces two metal pegs at a sufficent spacing, then talks at lo-fi with their surface team. Given the vagaries of local geology, mineralogical telluric currents and/or the aurora having a 'Bad Hair Day', results may vary...

 

[Merlin3189]

I'm thinking of 1-10 km cables in soil. Soil resistivity about 1000 ohms/meter. Cable resistivity about 0.1 ohms/meter. Soil to cable resistance ratio about 10⁴

But the cable diameter (maybe a few mm?) is much smaller than the diameter of the earth. Yes, I'm not assuming the whole Earth is involved, but once you move away from the ground rod itself, the current is traveling through an area of many sq m - a ratio of areas > 10⁶. What I suppose we should do is start with the surface area of the stake, caculate the resistance of a piece of subsoil dx thick, then sum a seies of such cylinders until we're a few m away. If you're pessimistic, perhaps use half of a cylinder. (I've tried this and still get unacceptably high values, so I'd be interested if anyone has found a link to a good treatment of this.)

The best description of this I've found so far is a leaflet by Fluke. There it is suggested that effective ground connections should be less than 25 Ω (NEC 250.56) or less than 5 Ω (NFPA and IEEE). This appears at odds with the typical values they give for grounding stakes in different soil types (which are in line with your 1 kΩm resistivity), except that they point out, a ground connection can be realized by an array of spaced ground stakes, plates or burried mesh, which can reduce the resistance by upto 40% for doubling the number of stakes. (I guess the 40% is a sort of cost benefit optimum, since to achieve 50% would require no interaction so very large spacing.)
So in good soil (<300 Ωm) a 16 stake array should meet the <5 Ω criterion. Dry sandy soil at 1000 Ωm might need over 100 stake array, or whatever the equivalent mesh or plates array is.) . Increasing the size depth of the conductors also helps.

Single line Earth return is used for power distribution in rural ares of NZ, Canada, Africa according to WikiP and others. Even at the multi kV levels and low currents (~10 A) used, I can't see their doing this if the power loss in the Earth return were very large. The only figure I can find for ground resistance here is 5 - 10 Ω, which is dramatically lower than anything I can calculate. It is stated that the economic benefit increases linearly with distance, which I don't think would be the case if the resistance increased with distance and resistive losses were significant. The effect of increasing wire resistance and its detriment on voltage regulation is mentioned, but not the Earth current. It may be that they include the Earth return resistance in with the resistance of the wire, but only materials and dimensions of the wire are mentioned.

3. RESISTANCE AND INDUCTANCE OF THE GROUND RETURN PATH According to Rudenberg(3), the effective resistance of the ground return current path is given by $$ R = π^{2}fl \times 10^{-7} Ω $$ where f = supply frequency /Hz, l = length of ground path/m
It should be noted that the resistance of the Earth path is dependent upon the frequency of current and length of path but is independent of Earth resistivity.

From Balwinder Singh Samra, "An analysis of single wire Earth return (SWER) system for rural electrification",Masters Thesis at University of Missouri-Rolla ,1972, referring to 3. Rudenberg, R., "Fundamental Consideration on Ground Currents," Electrical Engineering, Vol. 64, January 1945, 1-13.

This seems to give an Earth resistance of 0.05 Ω/km not including the electrode resistance at the ends, for which he gives these examples:

So this does not support my notion that the Earth has effectively zero resistance, but it does mean the Earth return resistance is dominated by the electrode resistance upto ~100 km. I can't find the Rudenberg info, obviously not as memorable as his electron microscope work, so I wonder if that, presumably approximate, formula is intended to be valid over larger distances?

 

[anorlunda]

I must admit that I never specifically studied single-phase-earth-return power lines. They are rare in the USA. So without that specific knowledge, I'm going to leave this conversation.

BTW, here is a picture of a medium voltage underground power cable.

 

[Merlin3189]

I must admit that I never specifically studied single-phase-earth-return power lines.

I neither. I hope I haven't given that impression. I merely sought confirmation, or otherwise, of what the situation seemed to me to be on a priori grounds. Having heard of SWER systems it seemed a sensible place to look.

 

[tech99]

I have an argument with some of my colleagues about the concept of ground path return. They say the electrical signal must return to the transmitter ground to complete the circuit. To me that is modelling the Earth as a resistor. That would certainly be true if we used a DC signal.

But we use an AC signal, typically 8 - 30 kHz, but sometimes as low as 500 Hz. I argue the Earth is better modeled as a capacitor, and the signal does not necessarily travel back to the ground stake.

We know that for DC the current definitely returns via the ground, as illustrated by the power feeding of the trans Atlantic communication cables, where a sea plate is used each end and the current finds its way back across the ocean. The ground current seems to follow the wire, as the charges are attracted to it. They do not wander all over the globe.
For higher frequencies we probably expect a TEM mode to travel down the system, like a coaxial cable. This would involve equal currents in ground and wire.
If the frequency is high and the wire is either some height above ground or in poor soil, there is the possibility of the single wire surface wave mode. This will deliver small amounts of power with just a local current sink at the receiving end (the system has high Zo so requires high voltages).. In addition, a surface wave will propagate over the ground and spread out in all directions until ground loses absorb the energy. We also expect to lose energy to free space radiation.
Earth has conductance and capacitive susceptance, and is thought to act primarily as a resistor up to about 10MHz and a lossy capacitor above that, but I cannot see exactly how that changes your mechanism.

 

[anorlunda]

This is all very interesting, and I'm learning stuff about things I never studied before (such as that transatlantic cable with sea plates, I would love a link for that.)

But we have to be careful with our words:

We know that for DC the current definitely returns via the ground,

Of course the most common case is that a wire provides the return path for DC or single phase AC. In previous threads, it has been mention that most wiring codes for electric power require a return conductor, thus excluding Earth returns. We have also had novices that mistakenly believe that all power wiring, in households and across continents use Earth return. That is not the case.

In this thread we are discussing things that are real but unusual.

 

[Tom.G]

From Balwinder Singh Samra, "An analysis of single wire Earth return (SWER) system for rural electrification",Masters Thesis at University of Missouri-Rolla ,1972, referring to 3. Rudenberg, R., "Fundamental Consideration on Ground Currents," Electrical Engineering, Vol. 64, January 1945, 1-13.

I can't find the Rudenberg info

R. Rudenberg was at Harvard at that time and Harvard has records of at least all alumni available online. To access the search https://alumni.harvard.edu/help/directory-search you must login with your Harvard ID/password. Any Harvard alumnus reading this care to put in a bit of time?

 

[tech99]

This is all very interesting, and I'm learning stuff about things I never studied before (such as that transatlantic cable with sea plates, I would love a link for that.)

This is a description of a modern power feeding equipment. https://www.nec.com/en/global/techrep/journal/g10/n01/pdf/100107.pdf.

 

[mcstatz5829]

To clarify from some posts above - most of the time the facility we are locating is insulated from the earth. And yes, it is indeed the magnetic field that we are detecting.

 

[mcstatz5829]

Earth has conductance and capacitive susceptance, and is thought to act primarily as a resistor up to about 10MHz and a lossy capacitor above that, but I cannot see exactly how that changes your mechanism.

So then at the frequencies we deal with, it acts primarily as a resistor. And from the SWER reference above, one with very low resistivity.

The practical question we have at hand is whether to train our technicians about ground path return, and if so, how. And that started the discussion if ground path return, in the form visualized as a DC signal (like our simplifications show it), is actually a thing at all.

 

[mcstatz5829]

I'm thinking of 1-10 km cables in soil. Soil resistivity about 1000 ohms/meter. Cable resistivity about 0.1 ohms/meter. Soil to cable resistance ratio about 10⁴

I don't understand this. You are talking about conductor to ground resistance? If so, the ohms/meter unit doesn't make much sense - it implies a 2 foot grounding stake would have double the resistance of a 1 foot grounding stakes, which doesn't make sense to me.

 

[Baluncore]

You have a circulating current. There are two ends to the circuit, one connected to either side of the AC generator. One circuit path is the insulated UG conductor that concentrates the current you will follow. The other is the deep volume of rock and salty groundwater that forms the wide “earth return” path under and around the region. That diffuse return path has a much smaller local field in the vicinity of the surface and the UG conductor. A 1km long UG conductor path on the surface will have return currents flowing over 1km to the left, to the right and 1km deep under the conductor.

If so, the ohms/meter unit doesn't make much sense - it implies a 2 foot grounding stake would have double the resistance of a 1 foot grounding stakes, which doesn't make sense to me.

You only need to drive the electrode to a depth where contact with damp soil is sufficient for signal current to flow. Most of the resistance in the return path will be where the electrode(s) contact the ground. If that Earth contact is too deep, or too good, the generator will struggle to produce enough current because the bulk of the Earth is a very good conductor with a huge cross-section. It is often the case with beginners that they drive the electrodes too deep, or add brine to improve the Earth contact, a practice that is usually only ever required in dry sand.