Substation earthing calculation



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.


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


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

Input data

\begin{equation} I_f = 25 kA \quad t_f = 0,6 s \quad \rho_e = 60 \Omega m \quad D = 77,3 m \end{equation}

\(I_k\) - 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


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

\begin{equation} A = \dfrac{I}{K} \sqrt{\dfrac{t_f}{ln\dfrac{\Theta_f+\beta}{\Theta_i+\beta}}} \end{equation}

\(A\) - 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
\begin{equation} A = \dfrac{25000}{78} \sqrt{\dfrac{0,6}{ln\dfrac{400+202}{30+202}}}=254,2 mm^2 \end{equation}


Supporting structures on substation are connected to lattice earthing by steel tape \(2 \cdot 40 \cdot 5 mm^2 = 400 mm^2\). \begin{equation} 254 mm^2 < 400 mm^2 \end{equation} The cross-section of steel tape is sufficient.