Modelling buried cables illuminated by incident fields for electromagnetic transient analysis in the Laplace domain

Abstract

In this work, a procedure to obtain two-port admittance models of buried cables illuminated by incident electromagnetic fields for the analysis of transients in the frequency domain is presented. The procedure is based on cascading two-port admittance models of various cable segments, avoiding the numerical problems that chain matrix representations may have. To obtain results in the time domain, the Numerical Laplace Transform method is used.

Appendices

Appendix

Azimuthal magnetic field

The return stroke model MTLE [26] is used in this paper, for this model the channel current is given by

$$i\left({y}^{{\prime}},t\right)=u(t-{y}^{{\prime}}/{v}_{f}) {e}^{-{y}^{{\prime}}/\lambda } i(0,t-{y}^{{\prime}}/{v}_{f})$$

(A.1)

where $u(t)$ is the Heaviside function, ${v}_{f}$ is the return stroke velocity, with a value of ${1.5\times 10}^{8} \mathrm{m}/\mathrm{s}$, $\lambda$ is the decay constant, which is considered equal to 2 km [26].

The Heidler function is used to represent the channel base current in (A.1), which is given by [25]:

$$i\left(0,t\right)=\frac{{I}_{1}}{{\eta }_{1}}{\left(\frac{t}{{\tau }_{11}}\right)}^{{n}_{1}}\frac{{e}^{-t/{\tau }_{12}}}{1+{\left(t/{\tau }_{11}\right)}^{{n}_{1}}}$$

(A.2)

where

$${\eta }_{1}=exp\left[-\left(\frac{{\tau }_{11}}{{\tau }_{12}}\right){\left(\frac{{n}_{1}{\tau }_{12}}{{\tau }_{11}}\right)}^{1/{n}_{1}}\right]$$

(A.3)

In addition $I_{1} = 28\,{\text{kA}}$, $\tau_{11} = 1.8\,\upmu {\text{s}}$, $\tau_{12} = 95\,\upmu {\text{s}}$ and $n_{1} = 2$.

According to [20], the channel base current can be represented with the function shown in Fig. 19 and defined by

$$i\left(t\right)= \frac{{I}_{0}}{T}\left[u(t)t-u(t-T)(t-T)\right]$$

(A.4)

The azimuthal magnetic field on the ground surface due to $i(t)$ is given by [20]

$$\begin{array}{c}{H}_{\phi p}(t,r,d=0)={H}_{S}\frac{{T}_{0}}{{v}_{R} T}\left[u\left(t-{T}_{0}\right)\left[\sqrt{1+{\left(\frac{{v}_{f} t}{r}\right)}^{2}-{{v}_{R}}^{2}}-1\right]\right.\\ \left. -u(t-{T}_{0}-T)\left[\sqrt{1+{\left(\frac{{v}_{f}(t-T)}{r}\right)}^{2}-{{v}_{R}}^{2}}-1\right]\right]\end{array}$$

(A.5)

where ${v}_{R}={v}_{f}/{c}_{0}$ is the relative velocity of the return stroke, ${c}_{0}$ is the speed of light, ${T}_{0}=r/{c}_{0}$ is the arrival time of the field wave at the observation point; ${H}_{S}$ is the magnetostatic field achieved after a long period of time given by

$${H}_{S}=\frac{{I}_{0}}{2\pi r}$$

(A.6)

where ${I}_{0}$ is the peak value of the channel current.

Fig. 19

Channel base current

The azimuthal magnetic field due to $i\left({y}^{{\prime}},t\right)$, or any arbitrary current, can be obtained by applying superposition. The time period of interest is divided into N intervals of duration Δt. Then a vector is generated with the current increments of the intervals, according to the following expression:

$$ \Delta i\left( {y^{\prime},j} \right) = i\left( {y^{\prime},j\Delta t} \right) - i\left( {y^{\prime },\left( {j - 1} \right)\Delta t} \right),\quad for\;j = 1,2, \ldots ,N $$

(A.7)

Expression (A.5) can be solved for a current of unit amplitude (${I}_{0}=1$) and a rise time $T=\Delta t$, in each of the N intervals. This results in a vector of magnetic fields ${H}_{I}(j)$. The total azimuthal magnetic field at $t=k{\Delta} {\mathrm {t}}$ $(1\le k\le N)$ is obtained with the superposition of the magnetic fields produced by k unit trapezoids, scaled by the current increments as follows

$${H}_{\phi p}\left(k\right)= \sum_{j=1}^{k}{H}_{I}\left(j\right)\Delta i\left({y}^{{\prime}},k-j+1\right)$$

(A.8)

The azimuthal magnetic field is transformed to the frequency domain and then used to obtain the horizontal electric field.