# Substation earthing calculation

## Introduction

Due to the fact that earthing resistance is influenced by many factors, there are no calculation methods, which give 100 % accuracy. There are many mathematical formulas, which are based on experience and theory.

## Objective

To choose proper cross-section of metallic steel tape for earthing of 110/30 kV substation.

## Standard

EN 50522:2010 Earthing of Power Installations Exceeding 1 kV A.C.

## Input data

$$I_f = 25 kA \quad t_f = 0,6 s \quad \rho_e = 60 \Omega m \quad D = 77,3 m$$

where:
$$I_k$$ - fault current
$$t_k$$ - fault time
$$\rho_e$$ - soil resistivity, taken from the standard EN 50522:2010 - table J.1 (see below)
$$D$$ - diameter of a circle, which area is the same as the lattice earthing

Soil resistivities for frequencies of alternating currents - range of values, which were frequently measured (table J.1 from EN 50522:2010)

 Type of soil Soil resistivity $$\rho_e [\Omega m]$$ Marshy soil 5 to 40 Loam, clay, humus 20 to 200 Sand 200 to 2 500 Gravel 2 000 to 3 000 Weathered rock mostly below 1 000 Sandstone 2 000 to 3 000 Granite up to 50 000 Moraine up to 30 000

## Calculations

Minimum cross-section of earthing conductor on substation can be calculated by:

$$A = \dfrac{I}{K} \sqrt{\dfrac{t_f}{ln\dfrac{\Theta_f+\beta}{\Theta_i+\beta}}}$$

where:
$$A$$ - cross-section, $$mm^2$$
$$I$$ - conductor current, A (RMS value)
$$t_f$$ - duration of fault current, s
$$K$$ - constant depending on the material of the current-carrying capacity, taken from table D.1 in EN 50522:2010(see below),$$A \cdot \sqrt{s}/mm^2$$
$$\beta$$ - reciprocal of the temperature coefficient of resistance of the current-carrying component at 0 $$^\circ C$$, EN 50522:2010(see below), $$^\circ C$$
$$O_i$$ - initial temperature, values may be taken from IEC 60287-3-1. If no value is laid down in the national tables, 20C as ambient ground temperature at a depth of 1 m should be adopted.
$$O_f$$ - final temperature, $$^\circ C$$

Material constants (table D.1 from EN 50522:2010)

 Material $$\beta [^\circ C]$$ $$K [A \cdot sqrt{s}/mm^2]$$ Copper 234,5 226 Aluminium 228 148 Steel 202 78
$$A = \dfrac{25000}{78} \sqrt{\dfrac{0,6}{ln\dfrac{400+202}{30+202}}}=254,2 mm^2$$

## Result

Supporting structures on substation are connected to lattice earthing by steel tape $$2 \cdot 40 \cdot 5 mm^2 = 400 mm^2$$. $$254 mm^2 < 400 mm^2$$ The cross-section of steel tape is sufficient.