Electrical resistance in the ground

[baltoche]

Dear All,

 

First, please apologise my bad English (I'm not from the UK...). If my wording is not clear enough, please don't hesitate to ask me.

 

I'm interested in dc current distribution into the ground for modelling purposes. I could read on a technical paper that the voltage between two points in a homogeneous ground (constant resistivity) could be calculated by V=Rho/(2xPixd)

 

where:

- Rho is the ground resistivity (in Ohm.m)

- V is the voltage between the two considered points (in V)

- d is the distance separating the two points.

 

The formula is extracted from the resistance of half a sphere and whose center is point A, whilst point B is on the surface of the half-sphere, the redus of such a sphere being d.

 

First there is a doubt on whether to consider half a sphere or a full sphere, but beyond the model itself, I, I find the formula to much simple... What do you think ?

 

Thanks and Regards.

[studiot]

This is called Coulomb's solution and is for a perfectly homogenous ground.

 

The hemisphere is because it assumes the charge source is at the surface. That is the test electrode is small compared to the distance through the ground.

So the ground is only half a sphere. The space above the ground is the other half.

The the electric potential is hemispherically distributed.

 

Resistivity surveying depends on finding divergences from this solution to indicate changes within the ground.

 

Note that usually two electrodes are not sufficient for the currents and voltages involved and that at least four terminal Wenner arrays are used.

 

http://gpg.geosci.xyz/en/latest/content/DC_resistivity/DC_principles.html

 

https://www.google.co.uk/search?hl=en-GB&source=hp&biw=&bih=&q=wenner+array&gbv=2&oq=wenner+array&gs_l=heirloom-hp.3..0l5.875.6016.0.6500.14.9.1.4.5.0.141.1016.2j7.9.0....0...1ac.1.34.heirloom-hp..1.13.1126.YtWo4Iq2TcY

[Danijel Gorupec]

Hi baltoche... can you please double-check the formula you mentioned: units do not match.

[baltoche]

Gents,

 

Thanks for your contributions.

@Daniel : you're right, instead of voltage, I meant the resistance...

@Studiot : you're also right, the hemisphere correspond to an earth resistance located at the surface of the ground.

 

Having said that, suppose two points both buried in the ground and separated by a distance d. Is it posible to define the resistance between them if the medium in which they are buried is homogeneous and has got a resistivity Rho ?

 

Thanks and good evening

To Studiot...

 

I went to the websites you kindly provided me with. Near the firsty one you quoted, I found an intersting site : http://gpg.geosci.xyz/en/latest/content/electromagnetics/principles_of_em_induction.html

 

If you go to this site, you canfind an interesting sketch :

http://gpg.geosci.xyz/en/latest/_images/Tx_Rx_schematic.jpg

where part of the signal sent to a transmitter to a receiver flows in the ground instead of using a nominal path at the surface of the ground.

This is exactly what I'm trying to do in dc ! I have several dc potentials (with respect to far earth) distributed at the top surface of the soil. Such potentials behave as current sources. For each of these dc currents sources, part of the current uses a nominal path to reach othe potentials on the surface, but another part uses the ground, including the far earth as a "secondary path". My goal is to modelise the soil between each of these ppoints as well as the far earth. Maybe I can try to draw something for you. Meanwhile, thanks again for your support.

 

Good night !

[studiot]

It is good that you are trying to tell us what you want to do.

 

But eddy currents at DC?

 

Can you provide a diagram for discussion?

[Enthalpy]

The formula isn't complete (even after the corrections) because the resistivity depends both on the electrode radius and on their distance.

 

If the distance clearly exceeds the diameter, you can model (approximately) the current density like the sum of two point sources at the center of the hemispherical electrodes. Then you write each current density as I/(2piR²), multiply by the resisitivity to get a field, and integrate from the plane of symmetry to the electrode's radius to get a half-voltage, from which a ratio gives a resistance.

 

When the distance clearly exceeds the electrode diameter, the resistance depends essentially on the electrode diameter, a radical drawback to detect deposits in the ground.

 

In a more general situation, perhaps you can offset the virtual point current sources within the hemispherical electrodes so that the voltage resulting from both sources is approximately uniform at the electrodes. This trick is perfect for the capacitance of bifilar lines because the equipotentials between parallel charged thin lines are cylinders, so you get an exact solution. Between points, it must be approximate.

 

The described setup is well-known (and refined) for oil (and water) prospection. Used by the Schlumberger brothers as they founded the company a century ago. Meanwhile little used, because acoustic ("seismic") prospection gives more detailed results and the equipment is mobile.

 

You should be aware that the ground's resistivity varies an awful lot depending on the ground water, and it depends radically on the frequency because water is very capacitive, which imposes strong limits on the frequency depending on what is measured and how.

 

People who measure a resistivity, not only of the ground, don't use two electrodes because of the contact resistance which is big, not reproducible, and masks the target's resistance. The very minimum for such a goal is a "four-point" scheme, where two electrodes wider apart inject a current and two electrodes closer to an other measure the resulting voltage. Compute as previously. Varying the distances make the setup sensitive to varied depths, so to some limited extent, you can reconstruct a profile of the resisitivity versus depth.

 

Because of electrochemical potentials, some sort of modulation is necessary at the injected current, on/off being the minimum.

[baltoche]

Thanks to Enthalpy for his reply, I do appreciate. For easiness reasons, I consider that my sol is uniform.

 

I've issued and attached a document inn which I tried to clarify my concerns? All your contributions will be welcome.

 

Kind ReSciences forum.pdfgards.

[Enthalpy]

If the wires are long as compared with the prospected depth, then the good model is not a hemisphere but a half (conductor A) or full (B, C) cylinder, with field as 1/R and potential as Log(). Don't forget the half or double current.

 

Two different conductivities make everything much more difficult. I'm not sure there is an algebraic solution. If the electrode C isn't parallel to the others, it also gets much worse.

 

On the diagram, the second electrode of the circuit that inject the current is missing. If this circuit injects current elsewhere - here into the soil - this current must come back to the circuit somewhere. The second electrode will change completely the shape of the injected current.

[baltoche]

To Enthalpy...

 

Thanks for your return. I agree with you on that the long wire A should be modelled by half a cylinder whilst B should be modelled by half a cylinder.

As a start, could we work with a common resistivity (rho). If you forget conductor C for the moment, are you able to model the ground with resistances ?

 

To answer your latest question, all currents leaving conductor A goes back to the negative of a rectifier somewhere

[Enthalpy]

I hope nobody feels excluded! Reducing a forum to a dialogue would be missing chances.

 

Baltoche, if you permit, I put here your diagram - ask the moderators to remove it if you don't want it.

 

If the conductors are clearly longer than the depth, consider they're infinitely long. Then a leaked current per unit length I/L gives at radius R a current density I/(L2piR) for the complete cylinder. The voltage gradient is rho times bigger, and a voltage difference is the integral, hence the resistance is Log(R_big/R_small)*rho/(L2pi).

 

This holds for one electrode only that injects current. From where the current returns, you get a similar voltage but with opposite current sign, so the position of the return contact is vital. And for the half-cylinder, double the resistance.

 

The voltage is finite only between two radii. One may take the electrode's radius in some cases, but the quality of the contact makes this doubtful. A better case is the already suggested four-point protocol, which works also with cylindric electrodes, and relieves the dependency on the contact resistance.

 

Are you from a French-speaking country?

[baltoche]

Hello Enthalpy,

 

You got me, I'm indeed from a French speaking country .

 

I also hope that nobody feels excluded. anyway, everybody is welcome to join us !

 

Yes, I can confirm that conductors are very long compared to their depth.

 

The current finally returns to conductor A, but as I divided this conductor into N sections, I've got approx. N nodes where this current can partly return. My real problem is to model my soil (both between nodes and between a given node and far earth) into a network of resistors, and to give a value to these resistors. It calls coupling notions that I don't have.

 

Thanks and Regards.

[Enthalpy]

If the conductor is good enough, maybe you don't need to cut it in sections. Sections would let you observe a potential difference along the length direction, which may be small and uninteresting. That would leave you with a two-dimensional problem, easier to solve.

 

Can you tell us more about the situation: is the current injection in the ground intentional? Is the conductor B a safety against current injection, a measure electrode...? Are you assessing the composition of the ground? Because, for each purpose, there are better electrode configurations, and these use to be easier to compute as well.

[baltoche]

My conductor is around 10 mOhm/km and my ground could be down to 30 Ohm.m. My conductor B corresponds to steel reinforcement and may be of the order of my conductor A resistance per length.

As far as my injection is concerned, it is not intentional. This currents comes from a transverse voltage between conductors A and B, knowing that the insulation resistance between A and B is far to be perfect, thus generating a current whose part reaches my ground.

My goal by sectionalising my conductor A is to get a network of resistors, that I'd like to resolve with Matlab.

 

I found a paper on Internet, slightly describing the environment of my model. I attached it for you. Section 2 is the most appropriate to my concern. The authors apply EMC model to a dc model in 3D arrangement. Section 2.2 is easy to understand. However, Section 2.3 (longitudinal model) is less easy for me. Nodes seem to be bonded to each other through a resistor but I cannot seize the formula. Maybe it is a classic formula for capacitive coupling on transmission lines in parallel, who knows... I'm far to be an EMC man and this may explain my lacks... Maybe it will more ring a bell to you.

 

 

Kind Regards.

 

PS : I have an issue in cutting and pasting text and figures on the forum. Is there a clue to achieve this ?

 

 

 

 

Startrack.pdf

[Enthalpy]

The button "more reply options" gives, among others, access to "attach files", which can be images.

 

So "conductor A" is a railway that brings current back to the supply, in DC or at 16.33 / 25 / 50Hz, and "conductor B" is a victim that runs parallel to it but deeper, for instance a pipeline, that may become corroded by the current.

 

This doesn't need the ridiculous fuss of the joined Pdf. Supposedly some contractor company or research group trying to impress the customer.

 

Evaluate the distance over which the rail and pipeline run parallel. Cut this length in 3 thirds, where the current plunges, runs in the victim, and emerges back to the rail.

 

The current and the resistance of the rail over 2/3 the distance (this is, between the middles of the first and third thirds) give you the harming voltage.

 

Evaluate the resistances of the 1st and 3rd thirds, from the resistivity of the ground, and with cylindrical and half-cylindrical models. To these two, add the resistance of the pipeline over 2/3 the distance, and the imperfect insulation between the rail and the ground - evaluate it as you can.

 

Divide the harming voltage by the computed resistance, and you know the current in the victim pipeline. Done.

 

Since the ground's resistivity and the rail's insulation are essentially unknown, this model is just as good as any other. You need no distributed network, no Matlab nor Spice - and consequently, it has chances of being correct.

[baltoche]

You've perfectly understood the context of my issue. I'm just adding that there is a conductor C (like a gas or water pipe, conductor B more being a slab steel reinforcement) that also behaves as a victim. Another variation is the soil resistivity along the line, as well as the rail insulation. This makes the circuit a bit more complex, hence the need of Matlab if I divide my line into short sections.

Having said that, I read carefully your model. What do you mean by "computed resistance" please ?

The ground resistivity and the rail insulation are assumed.

 

With these three conductors (A, B and C), is your model still applicable ?

 

Good night.

[Enthalpy]

Computed resistance: from cylindrical and half-cylindrical models.