Page 84 - Electrical Safety of Low Voltage Systems

The Earth 67

Example 4.2 A hemispherical electrode, embedded into a soil of resistivity 300
· m (e.g., sand–clay mixtures), has a radius of 2 m. The maximum fault current
I allowed by the protective device is 100 A. Determine:

1. The earth resistance of the electrode
2. The potential difference between the electrode and a point on the
earth’s surface at distance r = 10 m
Solution The earth resistance of the electrode is
300
R G = = = 23.8
2 r 0 4
Reference is made to Fig. 4.5.

I I I 1 1 300 × 100 1 1
V r 0 −r = − = − = − = 1911 V
2 r 0 2 r 2 r 0 r 2 2 10
Example 4.3 A hemispherical electrode of radius1mis embedded into a soil of
resistivity 600 · m (e.g., well-graded gravel). The center of the electrode is 14 m
away from an ECP. Determine the perspective touch voltage V ST and the touch
voltage V T a person may be subject to, upon the circulation of the ground-fault
current I = 25 A.
Solution
I I I 1 1 600 × 25 1 1
V ST = − = − = − = 2299 V
2 r 0 2 r 2 r 0 r 2 1 (14 + 1)
R BG = 2 = 1200

R B 1000
V T = V ST × = 2299 × = 1045 V.
R B + R BG 1000 + 1200
FAQs

Q. What is the relationship between the ground potential, as shown in the
curve in Fig. 4.5, and the equipotential surface in Fig. 4.7 for a hemispherical
electrode?
A. As shown in Fig. 4.5, by “entering” with a generic distance r from the
electrode, we can read the relative potential V r . Once this pair of coordinates is
known, we can determine the equipotential surface by drawing a hemisphere
of radius r to which we associate the ground potential V r . By doing this for
any distance r, we will obtain all the possible equipotential surfaces.
Q. Are a person’s feet two electrodes in parallel?
A. Human feet are conventionally modeled as two circular plates and can be
considered in parallel only if they are at least five times their radii, of length
10 cm, apart. We conventionally assume feet separated by 60 cm. In reality,
this might not always happen during faults. However, it is assumed that the