Earth ground resistance measurement

[cianfa72]

TL;DR

About the logic behind the Fall of potential method to measure the ground resistance

Starting from this Fall of potential method, earth ground resistance measurement of a ground electrode E driven into the ground can be done using the "Fall of potential method". Such a measurement method involves two auxiliary test probes (say S and H) driven into the ground at a certain distance form E.

As far as I understand, the logic behind the measurement of potential difference (ddp) between electrodes E and S by the voltmeter V boils down to following fact.

Assuming a semi-spherical model for the ground electrode E, at a linear distance R from it, a semi-sphere of diameter R of soil around it is involved. The current in the measurement circuit (forced by a current source attached between electrodes E and H) flows across the semi-spherical surface of radius R that defines an equipotential surface as well. The latter means an electrode S driven into the ground touching it will assume the potential of let's say "the ground terminal/side of the earth resistor ad distance R from where the electrode E has been driven into the ground".

Do you think it actually makes sense ?

 

[renormalize]

Do you think it actually makes sense ?

Can you provide a diagram of your electrodes, their connecting wires and your measuring devices?
Are you familiar with the theory of four-point probe resistance measurements?

 

[Baluncore]

Confusion arises in the first diagram, where equipotential lines cross the deep conductive electrode.

To measure ground resistance, independent of the electrode contact resistance, four equally spaced electrodes are normally used. The outer electrodes are used to insert a current into the ground, while the inner two electrodes are used for the voltage measurement. That eliminates the variable contact resistance of all electrodes from the equation. First study and understand that system.
https://en.wikipedia.org/wiki/Four-terminal_sensing

The four electrode system has been modified to the three electrode measurement, specifically to measure the ground contact resistance of one of the current electrodes.

 

[cianfa72]

To measure ground resistance, independent of the electrode contact resistance, four equally spaced electrodes are normally used. The outer electrodes are used to insert a current into the ground, while the inner two electrodes are used for the voltage measurement. That eliminates the variable contact resistance of all electrodes from the equation. First study and understand that system.
https://en.wikipedia.org/wiki/Four-terminal_sensing

Sorry, I was sloppy in describing my question.

Consider the following TT power distribution system diagram. $R_n$ is the ground resistance of the Neutral (N) grounded at the power grid facility. $R_t$ is the ground resistance at chassis equipment's point A where the PE conductor of the local domestic earthing system is attached to. $R_c$ and $R_{tc}$ are the resistance of the person touching point A (hand-to-feet) and its ground resistance respectively.

Now the point is: how ground resistances $R_n$ and $R_t$ are actually calculated ? I believe they also include the grounding electrode's contact resistance. Then, using for instance the "semi-spherical" model, one is able to calculate the "soil resistance" at a distance R from where the grounding electrode is driven into the earth.

To do that one consider basically a soil semi-sphere of radius R: the "terminals" of such "soil resistor" are the point where the electrode is grounded and the "semi-spherical boundary surface" of radius R.

 

[Baluncore]

To do that one consider basically a soil semi-sphere of radius R: the "terminals" of such "soil resistor" are the point where the electrode is grounded and the "semi-spherical boundary surface" of radius R.

"Semi-spherical"? No good can come from a word that is a misbegotten hybrid of two languages, Latin and Greek. Semicircle and hemisphere are acceptable compounds, semisphere and television are not.

The point where the electrode is grounded will have an infinite resistance, since a point has no surface area for conduction. The applied mathematics of a transition from a cylindrical electrode to a hemispherical shell is too difficult for me. I must assume that a real electrode is also hemispherical.

 

[cianfa72]

The point where the electrode is grounded will have an infinite resistance, since a point has no surface area for conduction. The applied mathematics of a transition from a cylindrical electrode to a hemispherical shell is too difficult for me. I must assume that a real electrode is also hemispherical.

Ok, so to simplify the math one can assume that the "soil hemispherical shell of radius R equivalent resistor" has two terminals or leads. The former is given by the grounding electrode assumed to be hemispherical too, the latter by the "radius R soil hemispherical shell boundary surface".

Makes sense ?

 

[Baluncore]

Makes sense ?

No, it does not.

Ok, so to simplify the math one can assume that the "soil hemispherical shell of radius R equivalent resistor" has two terminals or leads.

So there are two unspecified terminals.

The former is given by the grounding electrode assumed to be hemispherical too, the latter by the "radius R soil hemispherical shell boundary surface".

The terms 'former' and 'later', required that you earlier identify the two in order, but you did not do that. You may have known which you were referring to, but I'm not a mind reader.

Your post seems hollow, as if some critical statement is missing.

Electrodes are conductive, so should have the same profile as the equipotential where you expect them to connect.

 

[cianfa72]

The terms 'former' and 'later', required that you earlier identify the two in order, but you did not do that.

Perhaps I was unclear, take an hemispherical grounding rod/electrode of radius $r_0$ driven into Earth.

At a distance $R$ from the grounding point, the involved soil's resistance can be estimated by considering an hemispherical soil shell of external radius R and thickness $R - r_0$ with soil resistivity $\rho$. From a logical viewpoint what we have is a "hemispherical soil resistor" with 2 terminals/leads. The hemispherical electrode grounding resistance is therefore given by $$\int_{r_0}^{\infty} \frac {\rho} {2\pi r^2} \, dr$$
Basically, starting from a distance $\hat R$ from the electrode grounding point, the contribution of "hemispherical soil shells" to the total grounding resistance becomes reasonably negligible.

 

[renormalize]

Basically, starting from a distance $\hat R$ from the electrode grounding point, the contribution of "hemispherical soil shells" to the total grounding resistance becomes reasonably negligible.

@cianfa72, while that's true, you really don't need to worry about where the contribution becomes negligible since the grounding resistance is exactly calculable for your configuration. Start from your hemispherical grounding-conductor of radius $r_c$ that injects a current $I$ into a semi-infinite body of soil with uniform resistivity $\rho\,$:

The voltage field $V$ in the soil is found as follows. By the symmetry of the problem, any hemisphere in the soil that's centered on the conductor is an equipotential surface and so the electric field $\vec{E}$ on that surface is strictly radial and uniformly distributed:$$\vec{E}\left(R\right)=E\left(R\right)\hat{r}\tag{1}$$By Ohm's Law:$$E\left(R\right)=\rho J\left(R\right)\tag{2}$$the current density is consequently also radial and uniformly distributed, so conservation of the current requires:$$J\left(R\right)=\frac{I}{2\pi R^{2}}\tag{3}$$Combining eqs.(2) and (3) yields the magnitude of the electric field in the soil:$$E\left(R\right)=\frac{\rho I}{2\pi R^{2}}\tag{4}$$From this we can find the voltage-difference between the electrode and a point infinitely deep in the soil by integrating along any radius:$$\Delta V\left(r_{c}\right)\equiv V\left(r_{c}\right)-V\left(\infty\right)=-\intop_{\infty}^{r_{c}}E\left(R\right)dR=-\intop_{\infty}^{r_{c}}\frac{\rho I}{2\pi R^{2}}dR=\frac{\rho I}{2\pi r_{c}}\tag{5}$$The total grounding resistance $Z$ is therefore simply:$$Z\left(r_{c}\right)\equiv\frac{\Delta V\left(r_{c}\right)}{I}=\frac{\rho}{2\pi r_{c}}\tag{6}$$