Construction of a two-layer soil model from earth conductivity maps

Abstract

It is known that, when dealing with 50–60 Hz electromagnetic interference problems between power/railway lines and metallic pipelines/telecommunication cables, an important role is played by the soil modeling. To this aim, this paper presents a novel algorithm devoted to the construction of an equivalent two-layer soil model starting from the combined use of data available from different kinds of earth conductivity maps: on one side, the ones related to the geological characteristics relevant to the area of interest and, on the other side, the ones based on radio electric propagation measurements. From these data and by means of a suitable algorithm, the parameters characterizing the equivalent two-layer soil model are obtained.

Appendices

Appendix 1

The purpose of Appendix 1 is to give some information about the frequency dependence of the soil conductivity; Alipio and Visacro [17] on the basis of many measurements, proposed the following empirical formula that describes the soil conductivity σ(f) in function of the frequency in the range [100 Hz, 4 MHz]:

$$ \sigma \left( f \right) = \sigma_{0} \left\{ {1 + \left[ {1.2 \cdot 10^{ - 6} \left( {\frac{1}{{\sigma_{0} }}} \right)^{0.73} } \right]\left( {f - 100} \right)^{0.65} } \right\} $$

(10)

where σ(f) is expressed in S/m and σ₀ is the value of the conductivity at 100 Hz expressed in S/m. In Fig. 

Fig. 6

Ratio σ(f)/σ₀ versus frequency for different values of σ₀

6, we have plotted the value of the ratio σ(f)/σ₀ for different values of σ₀; we can notice that till to about 10 kHz the value of the ratio is always very close to 1 and increases more rapidly for lower values of σ₀ so meaning that soils which are poorly conductive at power frequencies become more conductive at 1 MHz even by a factor 2–3.

One could take into account of (10) in order to evaluate σ₁ and D when solving system (7). Table

Table 3 Comparison between some results with σ₂ constant and when derived from (10)

3 shows some results obtained when considering σ_geo constant and when considering it as function of the frequency according to (10) with σ_geo = σ₀. The quantities σ_1AV and D_AV are the first layer conductivity and thickness obtained by taking into account of (10).

By looking at Table 3, one can see that by considering σ₂ constant or variable in function of the frequency according to (10), has influence mainly on the result of the first layer thickness.

Appendix 2

The scope of Appendix 2 is to justify formula (4).

Starting from formula (2), we may write:

$$ {\text{e}}^{{ - 2D\sqrt {\pi f\mu_{0} \sigma_{1} } }} = \frac{{\left( {\sqrt {\sigma_{1} } + \sqrt {\sigma_{2} } } \right)\left( {\sqrt {\sigma_{1} } - \sqrt {\sigma_{e} \left( f \right)} } \right)}}{{\left( {\sqrt {\sigma_{1} } - \sqrt {\sigma_{2} } } \right)\left( {\sqrt {\sigma_{1} } + \sqrt {\sigma_{e} \left( f \right)} } \right)}} $$

(11)

The left-hand side of (11) represents a real and positive number, so the following inequality must hold:

$$ \frac{{\left( {\sqrt {\sigma_{1} } + \sqrt {\sigma_{2} } } \right)\left( {\sqrt {\sigma_{1} } - \sqrt {\sigma_{e} \left( f \right)} } \right)}}{{\left( {\sqrt {\sigma_{1} } - \sqrt {\sigma_{2} } } \right)\left( {\sqrt {\sigma_{1} } + \sqrt {\sigma_{e} \left( f \right)} } \right)}} > 0 $$

(12)

Moreover, in (3) that represents a real positive quantity, the argument of the logarithm must be greater than 1; thus, it has to be:

$$ \frac{{\left( {\sqrt {\sigma_{1} } - \sqrt {\sigma_{2} } } \right)\left( {\sqrt {\sigma_{1} } + \sqrt {\sigma_{e} \left( f \right)} } \right)}}{{\left( {\sqrt {\sigma_{1} } + \sqrt {\sigma_{2} } } \right)\left( {\sqrt {\sigma_{1} } - \sqrt {\sigma_{e} \left( f \right)} } \right)}} > 1 $$

(13)

By combining together (12) and (13), one finally obtains the following inequality:

$$ {0 < }\frac{{\left( {\sqrt {\sigma_{1} } + \sqrt {\sigma_{2} } } \right)\left( {\sqrt {\sigma_{1} } - \sqrt {\sigma_{e} \left( f \right)} } \right)}}{{\left( {\sqrt {\sigma_{1} } - \sqrt {\sigma_{2} } } \right)\left( {\sqrt {\sigma_{1} } + \sqrt {\sigma_{e} \left( f \right)} } \right)}}{ < 1} $$

(14)

In order to solve (14), we have to distinguish two cases:

Case 1: σ₂ < σ_e(f).

One can verify that (14) is fulfilled for σ₁ > σ_e(f).

Case 2: σ₂ > σ_e(f).

One can verify (14) is fulfilled for σ₁ < σ_e(f).

Thus, one has the relationships in (4).

Appendix 3

Appendix 3 is devoted to a concise demonstration of conditions (8a) and (8b) able to guarantee the existence of the solution of system (7).

For convenience, let us introduce the expressions that represent the first layer thickness in function of σ₁, σ₂ and of the measured equivalent conductivity σ_e at 1 MHz and 10 kHz, respectively:

$$ D_{{1{\text{MHz}}}} \left( {\sigma_{1} } \right) = \frac{{\ln \left[ {\frac{{\left( {\sqrt {\sigma_{1} } - \sqrt {\sigma_{2} } } \right)\left( {\sqrt {\sigma_{1} } + \sqrt {\sigma_{e} \left( {10^{6} } \right)} } \right)}}{{\left( {\sqrt {\sigma_{1} } + \sqrt {\sigma_{2} } } \right)\left( {\sqrt {\sigma_{1} } - \sqrt {\sigma_{e} \left( {10^{6} } \right)} } \right)}}} \right]}}{{2\sqrt {\pi {\kern 1pt} 10^{6} \mu_{0} \sigma_{1} } }} $$

(15)

$$ D_{{10{\text{kHz}}}} \left( {\sigma_{1} } \right) = \frac{{\ln \left[ {\frac{{\left( {\sqrt {\sigma_{1} } - \sqrt {\sigma_{2} } } \right)\left( {\sqrt {\sigma_{1} } + \sqrt {\sigma_{e} \left( {10^{4} } \right)} } \right)}}{{\left( {\sqrt {\sigma_{1} } + \sqrt {\sigma_{2} } } \right)\left( {\sqrt {\sigma_{1} } - \sqrt {\sigma_{e} \left( {10^{4} } \right)} } \right)}}} \right]}}{{2\sqrt {\pi {\kern 1pt} 10^{4} \mu_{0} \sigma_{1} } }} $$

(16)

The solution of system (7) is given by the intersection of the curves (15) and (16) and is represented by the couple (σ₁, D) where both σ₁ and D must be positive.

Having in mind (4), in order that both the curves (15) and (16) are positive, one of the two following conditions must be fulfilled:

• I. σ₂ < min(σ_e1MHz, σ_e10kHz) AND σ₁ > max(σ_e1MHz, σ_e10kHz)

or

• II. σ₂ > max(σ_e1MHz, σ_e10kHz) AND σ₁ < min(σ_e1MHz, σ_e10kHz)

Nevertheless, these two conditions are necessary but not sufficient in order to have the intersection of the curves (15) and (16); in fact, we have the following sub-cases:

Case Ia: σ₂ < σ_e1MHz < σ_e10kHz.

One can verify that it is always D_10kHz(σ₁) > D_1MHz(σ₁) for any value of σ₁; for σ₁ → ∞ both the curves tend to 0. In this case no intersection exists.

Case Ib: σ₂ < σ_e10kHz < σ_e1MHz.

Let us put into evidence the following points:

• the curve D_1MHz(σ₁) has a vertical asymptote for σ₁ = σ_e1MHz while curve D_10kHz(σ₁) has a vertical asymptote for σ₁ = σ_e10kHz; being σ_e10kHz < σ_e1MHz, one has that in a very small neighborhood at the right of σ_e1MHz we have that D_1MHz(σ₁) → ∞ while D_10kHz(σ₁) has a finite value; this means that D_1MHz(σ₁) > D_10kHz(σ₁)

• the following the relation holds:

  $$ \mathop {\lim }\limits_{{\sigma_{1} \to \;\infty }} \quad \frac{{D_{1{\rm MHz}} \left( {\sigma_{1} } \right)}}{{D_{10{\rm kHz}} \left( {\sigma_{1} } \right)}} = \frac{{\sqrt {\sigma_{2} } - \sqrt {\sigma_{e1{\rm MHz}} } }}{{\sqrt {\sigma_{2} } - \sqrt {\sigma_{e10{\rm kHz}} } }}\sqrt {\frac{{f_{2} }}{{f_{1} }}} $$

  (17)

If the above limit given by (17) is < 1, it means that it exists a particular value σ₁^* such that D_1MHz(σ₁) < D_10kHz(σ₁) for σ₁ > σ₁^*. By taking into account of the first item, we can conclude that: in the interval (σ_e1MHz, σ₁^*), one has D_1MHz(σ₁) > D_10kHz(σ₁) while in the interval (σ₁^*, ∞), one has D_1MHz(σ₁) < D_10kHz(σ₁).

This implies that the two curves have an intersection; thus, system (7) has solution provided that condition (8b) (which derives from (17) and from (9) with σ₂ = σ_geo) is fulfilled.

Case IIa: σ₂ > σ_e1MHz > σ_e10kHz.

One can verify that always D_10kHz(σ₁) > D_1MHz(σ₁) for any value of σ₁; for σ₁ → 0⁺ both the curves tend to the horizontal asymptote given by (5) with D_asym(10⁴) > D_asym(10⁶). In this case no intersection exists.

Case IIb: σ₂ > σ_e10kHz > σ_e1MHz.

Let us put into evidence the following points:

• The curve D_1MHz(σ₁) has a vertical asymptote for σ₁ = σ_e1MHz while curve D_10kHz(σ₁) has a vertical asymptote for σ₁ = σ_e10kHz; being σ_e1MHz < σ_e10kHz, one has that in a very small neighborhood at the left of σ_e1MHz we have that D_1MHz(σ₁) → ∞ while D_10kHz(σ₁) has a finite value; this means that D_1MHz(σ₁) > D_10kHz(σ₁).

• The following the relation holds:

  $$ \mathop {\lim }\limits_{{\sigma_{1} \to \;0^{ + } }} \quad \frac{{D_{10{\rm kHz}} \left( {\sigma_{1} } \right)}}{{D_{1{\rm MHz}} \left( {\sigma_{1} } \right)}} = \frac{{\sqrt {\sigma_{2} } - \sqrt {\sigma_{e10{\rm kHz}} } }}{{\sqrt {\sigma_{2} } - \sqrt {\sigma_{e1{\rm MHz}} } }}\sqrt {\frac{{f_{1} }}{{f_{2} }}} \sqrt {\frac{{\sigma_{e1{\rm MHz}} }}{{\sigma_{e10{\rm kHz}} }}} $$

  (18)

If the above limit given by (18) is > 1 it means that it exists a particular value σ₁^* such that D_1MHz(σ₁) < D_10kHz(σ₁) for σ₁ < σ₁^*. By taking into account of the first item, we can conclude that: in the interval (σ₁^*, σ_e1MHz), one has D_1MHz(σ₁) > D_10kHz(σ₁) while in the interval (0, σ₁^*), one has D_1MHz(σ₁) < D_10kHz(σ₁).

This implies that the two curves have an intersection; thus, system (7) has solution provided that condition (8a) (which derives from (18) and from (9) with σ₂ = σ_geo) is fulfilled.